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r 

UNIVERSITY -OF CALIFORNIA 
AT LOS ANGELES 





INTKODUCTION 



TO 



QUATERNIONS, 



WITH NUMEROUS EXAMPLES. 



INTRODUCTION 



TO 



QUATEENIONS, 



WITH NUMEEOUS EXAMPLES. 



BY 



P. KELLAND, M.A, F.R.S, 

FORMERLY FELLOW OF QUEENS* COLLEGE, CAMBRIDGE ; 
AND 

P. G. TAIT, M.A., SEC. R.S.E., 

FORMERLY FELLOW OF ST PETER'S COLLEGE, CAMBRIDGE J 

PROFESSORS IN THE DEPARTMENT OF MATHEMATICS IN THE 
UNIVERSITY OF EDINBURGH. 



SECOND EDITION: ...,*; 



Honlron : 
MACMILLAN AND CO. 

1882 

[All Eights reserved.] 



FEINTED BY C. J. CLAY, M.A. 

AT THE UNIVERSITY PRESS. 



Mitkemstictl 
Sciences 
Library 



K28i 



PREFACE. 

Ix preparing this second edition for press I have altered as 
slightly as possible those portions of the work which were 
written entirely by Prof. Kelland. The mode of presentation 
which he employed must always be of great interest, if only 
from the fact that he was an exceptionally able teacher ; but 
the success of the work, as an introduction to a method which 
is now rapidly advancing in general estimation, would of itself 
have been a sufficient motive for my refraining from any 
serious alteration. 

A third reason, had such been necessary, would have pre- 
sented itself in the fact that I have never considered with the 
necessary care those metaphysical questions connected with 
the growth and development of mathematical ideas, to which 
my late venerated teacher paid such particular attention. 

My own part of the book (including mainly Chap. X. and 
worked out Examples 10 24 in Chap. IX.) was written 
hurriedly, and while I was deeply engaged with work of a very 
different kind ; so that I had no hesitation in determining to 
re-cast it where I fancied I could improve it. 

P. G. TAIT. 



UNIVERSITY OP EDINBUEOH, 
November, 1881. 



210949 

274 



PEEFACE TO THE FIRST EDITION. 

THE present Treatise is, as the title-page indicates, the joint 
production of Prof. Tait and myself. The preface I write 
in the first person, as this enables me to offer some personal 
explanations. 

For many years past I have been accustomed, no doubt 
very imperfectly, to introduce to my class the subject of 
Quaternions as part of elementary Algebra, more with the 
view of establishing principles than of applying processes. 
Experience has taught me that to induce a student to think 
for himself there is nothing so effectual as to lay before him 
the different stages of the development of a science in some- 
thing like the historical order. And justice alike to the 
student and the subject forbade that I should stop short at 
that point where, more simply and more effectually than at 
any other, the intimate connexion between principles and pro- 
cesses is made manifest. Moreover, in lecturing on the ground- 
work on which the mathematical sciences are based, I could 
not but bring before my class the names of great men who 
spoke in other tongues and belonged to other nationalities 
than their own Diophantus, Des Cartes, Lagrange, for in- 
stance and it was not just to omit the name of one as 



Vlll PREFACE. 

great as any of them, Sir William Kowan Hamilton, who 
spoke their own tongue and claimed their own nationality. 
It is true the name of Hamilton has not had the impress 
of time to stamp it with the seal of immortality. And it 
must be admitted that a cautious policy which forbids to 
wander from the beaten paths, and encourages converse 
with the past rather than interference with the present, is 
the true policy of a teacher. But in the case before us, 
quite irrespective of the nationality of the inventor, there 
is ample ground for introducing this subject of Quaternions 
into an elementary course of mathematics. It belongs to 
first principles and is their crowning and completion. It 
brings those principles face to face with operations, and thus 
not only satisfies the student of the mutual dependence of 
the two, but tends to carry him back to a clear apprehension 
of what he had probably failed to appreciate in the sub- 
ordinate sciences. 

Besides, there is no branch of mathematics in which 
results of such wide variety are deduced by one uniform 
process; there is no territory like this to be attacked 
and subjugated by a single weapon. And what is of the 
utmost importance in an. educational point of view, the 
reader of this subject does not require to encumber his 
memory with a host of conclusions already arrived at in 
order to advance. Every problem is more or less self- 
contained. This is my apology for the present treatise. 

The work is, as I have said, the joint production 
of Prof. Tait and myself. The preface I have written 
without consulting my colleague, as I am thus enabled 



PREFACE. ix 

to say what could not otherwise have been said, that 
mathematicians owe a lasting debt of gratitude to Prof. 
Tait for the singleness of purpose and the self-denying 
zeal with which he has worked out the designs of his 
friend Sir Wm. Hamilton, preferring always the claims of 
the science and of its founder to the assertion of his own 
power and originality in its development. For rny own 
part I must confess that my knowledge of Quaternions 
is due exclusively to him. The first work of Sir Wm. 
Hamilton, Lectures on Quaternions, was very dimly and im- 
perfectly understood by me and I dare say by others, until 
Prof. Tait published his papers on the subject in the 
Messenger of Mathematics. Then, and not till then, did 
the science in all its simplicity develope itself to me. Sub- 
sequently Prof. Tait has published a work of great value 
and originality, An Elementary Treatise on Quaternions. 

The literature of the subject is completed in all but 
what relates to its physical applications, when I mention in 
addition Hamilton's second great work, Elements of Quater- 
nions, a posthumous work so far as publication is concerned, 
but one of which the sheets had been corrected by the 
author, and which bears all the impress of his genius. But 
it is far from elementary, whatever its title may seem to 
imply; nor is the work of Prof. Tait altogether free from 
difficulties. Hamilton and Tait write for mathematicians, 
and they do well, but the time has come when it behoves 
some one to write for those who desire to become mathe- 
maticians. Friends and pupils have urged me to undertake 
this duty, and after consultation with Prof. Tait, who from 



X PREFACE. 

being my pupil in youth is my teacher in riper years, 
I have, in conjunction with him, and drawing unreservedly 
from his writings, endeavoured in the first nine chapters 
of this treatise to illustrate and enforce the principles of 
this beautiful science. The last chapter, which may be 
regarded as an introduction to the application of Quater- 
nions to the region beyond that of pure geometry, is due 
to Prof. Tait alone. Sir W. Hamilton, on nearly the last 
completed page of his last work, indicated Prof. Tait as 
eminently fitted to carry on happily and usefully the appli- 
cations, mathematical and physical, of Quaternions, and as 
likely to become in the science one of the chief successors 
of its inventor. With how great justice, the reader of this 
chapter and of Prof. Tait's other writings on the subject 
will judge. 

PHILIP KELLAND. 

UNIVEESITT OF EDINBURGH, 
October, 1873. 



CONTENTS. 



CHAPTER L 

PAGES 

INTRODUCTORY 1 5 



CHAPTER II. 
VECTOR ADDITION AND SUBTRACTION 6 31 

Definition of a VECTOR, with conclusions immediately resulting 
therefrom, Art. 1 6 ; examples, 7 ; definition of UNIT VECTOR and 
TENSOR, with examples, 8; coplanarity of three coinitial vectors, 
with conditions requisite for their terminating in a straight line, 
and examples, 9 13 ; mean point, 14. 

ADDITIONAL EXAMPLES TO CHAPTER IL 



CHAPTER m. 
VECTOR MULTIPLICATION AND DIVISION 32 68 

Definition of multiplication, and first principles, Art. 15 18 ; 
fundamental theorems of multiplication, 19 22 ; examples, 23; 
definitions of DIVISION, VERSOR and QUATERNION, 24 28 ; examples, 
29 ; conjugate quaternions, 30 ; interpretation of formulae, 31. 

ADDITIONAL EXAMPLES TO CHAPTER IIL 



Xll CONTENTS. 

CHAPTER IV. 

PAGES 

THE STRAIGHT LINE AND PLANE 59 72 

Equations of a straight line and plane, 32, 33 ; modifications and 
results length of perpendicular on a plane condition that four 
points shall lie in the same plane, &c. 34; examples, 35. 

ADDITIONAL EXAMPLES TO CHAPTER IV. 



CHAPTER V. 
THE CIRCLE AND SPHERE . . 7390 

Equations of the circle, -with examples, 36, 37; tangent to circle 
and chord of contact, 38, 39; examples, 40 ; equations of the sphere, 
with examples, 41, 42. 

ADDITIONAL EXAMPLES TO CHAPTER V. 



CHAPTER VI. 
THE ELLIPSE . . . . . 91105 

Equations of the ellipse, 43 ; properties of <pp, 44 ; equation of 
tangent, 45; Cartesian equations, 46; (j>~ l p, ^p, &c. 47; properties 
of the ellipse, -with examples, 48 50. 

ADDITIONAL EXAMPLES TO CHAPTER VI. 



CHAPTER VII. 

THE PARABOLA AND HYPERBOLA 106 127 

-\ 

Equation of the parabola in terms of <j>p, with examples, 52 54; 
equations of the parabola, ellipse and hyperbola in a form corre- 
sponding to those with Cartesian co-ordinates, with examples, 55. 

ADDITIONAL EXAMPLES TO CHAPTER VII. 



CONTENTS. xiii 

CHAPTEE VIII. 

PAGES 

CENTRAL SURFACES OF THE SECOND ORDER 128 153 

Equation of the ellipsoid, 56; tangent plane and perpendicular 
on it, 57, 58; polar plane, 59, 60; conjugate diameters and diame- 
tral planes, with examples, 60 64 ; the cone, 65, 66 ; examples on 
central surfaces, 67 ; Pascal's hexagram, 68 

ADDITIONAL EXAMPLES TO CHAPTER VIII. 



CHAPTEE IX. 

FORMULAE AND THEIR APPLICATION 154 181 

Formulas, 69, 70; examples, 71. 

ADDITIONAL EXAMPLES TO CHAPTER IX. 

CHAPTEE X. 

VECTOR EQUATIONS OF THE FIRST DEGREE 182 212 

APPENDIX . 213232 



INTRODUCTION TO QUATERNIONS. 



CHAPTER I. 

INTRODUCTORY. 

THE science named Quaternions by its illustrious founder, Sir 
William Rowan Hamilton, is the last and the most beautiful ex- 
ample of extension by the removal of limitations. 

The Algebraic sciences are based on ordinary arithmetic, start- 
ing at first with all its restrictions, but gradually freeing themselves 
from one and another, until the parent science scarce recognises 
itself in its offspring. A student will best get an idea of the thing 
by considering one case of extension within the science of Arith- 
metic itself. There are two distinct bases of operation in that 
science addition and multiplication. In the infancy of the science 
the latter was a mere repetition of the former. Multiplication was, 
in fact, an abbreviated form of equal additions. It is in this form 
that it occurs in the earliest writer on arithmetic whose works have 
come down to us Euclid. Within the limits to which his prin- 
ciples extended, the reasonings and conclusions of Euclid in his 
seventh and following Books are absolutely perfect. The demon- 
stration of the rule for finding the greatest common measure of 
two numbers in Prop. 2, Book VII. is identically the same as that 
which is given in all modern treatises. But Euclid dares not 
venture on fractions. Their properties were probably all but un- 
known to him. Accordingly we look in vain for any demonstration 
of the properties of fractions in the writings of the Greek arith- 
meticians. For that we must come lower down. On the revival 
T. Q. 1 



2 QUATERNIONS. [CHAP. 

of science in the West, we are presented with categorical treatises 
on arithmetic. The first printed treatise is that of Lucas de Burgo 
in 1494. The author considers a fraction to be a quotient, and 
thus, as>he expressly states, the order of operations becomes the 
reverse of that for whole numbers multiplication precedes addi- 
tion, etc. In our own country we have a tolerably eai'ly writer on 
arithmetic, Robert Record, who dedicated his work to King Edward 
the Sixth. The ingenious author exhibits his treatise in the form 
of a dialogue between master and scholar. The scholar battles 
long with this difficulty that multiplying a thing should make it 
less. At first, the master attempts to explain the anomaly by 
reference to proportion, thus : that the product by a fraction bears 
the same proportion to the thing multiplied that the multiplying 
fraction does to unity. The scholar is not satisfied ; and accord- 
ingly the master goes on to say : "If I multiply by more than one, 
the thing is increased ; if I take it but once, it is not changed; and 
if I take it less than once, it cannot be so much as it was before. 
Then, seeing that a fraction is less than one, if I multiply by a 
fraction, it follows that I do take it less than once", etc. The 
scholar thereupon replies, " Sir, I do thank yon much for this 
reason ; and I trust that I do perceive the thing". 

Need we add that the same difficulty which the scholar in the 
time of King Edward experienced, is experienced by every thinking 
boy of our own times; and the explanation afforded him is precisely 
the same admixture of multiplication, proportion, and division which 
suggested itself to old Robert Record. Every schoolboy feels that 
to multiply by a fraction is not to multiply at all in the sense in 
which multiplication was originally presented to him, viz. as an 
abbreviation of equal additions, or of repetitions of the thing multi- 
plied. A totally new view of the process of multiplication has 
insensibly crept in by the advance from whole numbers to fractions. 
So new, so different is it, that we are satisfied Euclid in his logical 
and unbending march could never have attained to it. It is only 
by standing loose for a time to logical accuracy that extensions in 
the abstract sciences extensions at any rate which stretch from 
one science to another are effected. Thus Diophantus in his 



I.] INTRODUCTORY. 3 

Treatise on Arithmetic (i.e. Arithmetic extended to Algebra) 
boldly lays it down as a definition or first principle of his science 
that 'minus into minus makes plus'. The science he is founding 

* O 

is subject to this condition, and the results must be interpreted 
consistently with it. So far as this condition does not belong to 
ordinary arithmetic, so far the science extends beyond ordinary 
arithmetic : and this is the distance to which it extends It makes 
subtraction to staud by itself, apart from addition; or, at any rate, 
not dependent on it. 

"We trust, then, it begins to be seen that sciences are extended 
by the removal of barriers, of limitations, of conditions, on which 
sometimes their very existence appears to depend. Fractional 
arithmetic was an impossibility so long as multiplication was re- 
garded as abbreviated addition ; the moment an extended idea was 
entertained, ever so illogically, that moment fractional arithmetic 
started into existence. Algebra, except as mere symbolized arith- 
metic, was an impossibility so long as the thought of subtraction 
was chained to the requirement of something adequate to subtract 
from. The moment Diophantus gave it a separate existence 
boldly and logically as it happened by exhibiting the law of minus 
in the forefront as the primary definition of his science, that moment 
algebra in its highest form became a possibility ; and indeed the 
foundation-stone was no sooner laid than a goodly building arose 
on it. 

The examples we have given, perhaps from their very simplicity, 
escape notice, but they are not less really examples of extension 
from science to science by the removal of a restriction. We have 
selected them in preference to the more familiar one of the extension 
of the meaning of an index, whereby it becomes a logarithm, because 
they prepare the way for a further extension in the same direction 
to which we are presently to advance. Observe, then, that in frac- 
tions and in the rule of signs, addition (or subtraction) is very 
slenderly connected with multiplication (or division). Arithmetic 
as Euclid left it stands on one support, addition only, inasmuch 
as with him multiplication is but abbreviated addition. Arithmetic 
in its extended form rests on two supports, addition and multiplica- 

12 



4 QUATERNIONS. [CHAP. 

tion, the one different from the other. This is the first idea we 
want our reader to get a firm hold of ; that multiplication is not 
necessarily addition, but an operation self-contained, self-interpret- 
able springing originally out of addition ; but, when full-grown, 
existing apart from its parent. 

The second idea we want our reader to fix his mind on is this, 
that when a science has been extended into a new form, certain 
limitations, which appeared to be of the nature of essential truths 
in the old science, are found to be utterly untenable ; that it is, in 
fact, by throwing these limitations aside that room is made for the 
growth of the new science. We have instanced Algebra as a growth 
out of Arithmetic by the removal of the restriction that subtraction 
shall require something to subtract from. The word 'subtraction' 
may indeed be inappropriate, as the word multiplication ap- 
peared to be to Record's scholar, who failed to see how the multi- 
plication of a thing could make it less. In the advance of the 
sciences the old terminology often becomes inappropriate ; but if 
the mind can extract the right idea from the sound or sight of a 
word, it is the part of wisdom to retain it. And so all the old words 
have been retained in the science of Quaternions to which we are 
now to advance. 

The fundamental idea on which the science is based is that of 
motion of transference. Real motion is indeed not needed, any 
more than real superposition is needed in Euclid's Geometry. An 
appeal is made to mental ti'ansference in the one science, to mental 
superposition in the other. 

We are then to consider how it is possible to frame a new science 
which shall spring out of Arithmetic, Algebra, and 'Geometry, and 
shall add to them, the idea of motion of transference. It must be 
confessed the project we entertain is not a project due to the 
nineteenth century. The Geometry of Des Cartes was based on 
something very much resembling the idea of motion, and so far the 
mere introduction of the idea of transference was not of much value. 
The real advance was due to the thought of severing multiplication 
from addition, so that the one might be the representative of a kind 
of motion absolutely different from that which was represented by 



I.] INTRODUCTORY. 5 

the other, yet capable of being combined with it. What the nine- 
teenth century has done, then, is to divorce addition from multipli- 
cation in the new form in which the two are presented, and to 
cause the one, in this new character, to signify motion forwards 
and backwards, the other motion round and round. 

We do not purpose to give a history of the science, and shall 
accordingly content ourselves with saying, that the notion of sepa- 
rating addition from multiplication attributing to the one, motion 
from a point, to the other motion about a point had been floating 
iu the minds of mathematicians for half a century, without producing 
many results worth recording, when the subject fell into the hands 
of a giant, Sir William Rowan Hamilton, who early found that his 
road was obstructed- he knew not by what obstacle so that many 
points which seemed within his reach were really inaccessible. He 
had done a considerable amount of good work, obstructed as he was, 
when, about the year 1843, he perceived clearly the obstruction to 
his progress in the shape of an old law which, prior to that time, 
had appeared like a law of common sense. The law in question is 
known as the commutative law of multiplication. Presented in its 
simplest form it is nothing more than this, ' five times three is the 
same as three times five'; more generally, it appears under the 
form of 'ab = ba whatever a and b may represent'. When it 
came distinctly into the mind of Hamilton that this law is not a 
necessity, with the extended signification of multiplication, he saw 
his way clear, and gave up the law. The barrier being removed, 
he entered on the new science as a warrior enters a besieged city 
through a practicable breach. The reader will find it easy to enter 
after him. 



CHAPTER II. 

VECTOR ADDITION AND SUBTRACTION. 

1. Definition of a Vector. A vector is the representative of 
transference through a given distance, in a given direction. Thus 
if AB be a straight line, the idea to be attached to 'vector AB' is 
that of transference from A to B. 

For the sake of definiteness we shall frequently abbreviate the 
phrase ' vector AB ' by a Greek letter, retaining in the meantime 
(with one exception to be noted in the next chapter) the English 
letters to denote ordinary numerical quantities. 

If we now start from .Band advance to (7 in the same direction, 
BC being equal to AB, we may, as in ordinary geometry, designate 
' vector EG ' by the same symbol, which we adopted to designate 
' vector AB.' 

Further, if we start from any other point in space, and 
advance from that point by the distance OX equal to and in the 
same direction as AB, we are at liberty to designate 'vector OX' 
by the same symbol as that which represents AB. 

Other circumstances will determine the starting point, and in- 
dividualize the line to which a specific vector corresponds. Our 
definition is therefore subject to the following condition : All lines 
which are equal and drawn in the same direction are represented by 
the same vector symbol. 

We have purposely employed the phrase 'drawn in the same 
direction ' instead of ' parallel,' because we wish to guard the 
student against confounding 'vector AB ' with 'vector BA.' 



ART. 2.] VECTOR ADDITION AND SUBTRACTION. 7 

2. In order to apply algebra to geometry, it is necessary to 
impose on geometry the condition that when a line measured in 
one direction is represented by a positive symbol, the same line 
measured in the opposite direction must be represented by the cor- 
responding negative symbol. 

In the science before us the same condition is equally requisite, 
and indeed the reason for it is even more manifest. For if a 
transference from A to B be represented by + a, the transference 
which neutralizes this, and brings us back again to A, cannot be 
conceived to be represented by anything but a, provided the 
symbols + and are to retain any of their old algebraic meaning. 
The vector AB, then, being represented by + a, the vector BA will 
be represented by - a. 

3. Further it is abundantly evident that so far as addition and 
subtraction of parallel vectors are concerned, all the laws of Algebra 
must be applicable. Thus (in Art. 1) AB + BC or a + a produces 
the same result as AC which is twice as great as AB, and is there- 
fore properly represented by 2a ; and so on for all the rest. The 
distributive law of addition may then be assumed to hold in all its 
integrity so long at least as we deal with vectors which are paralk-1 
to one another. In fact there is no reason whatever, so far, why 
a should not be treated in every respect as if it were an ordinary 
algebraic quantity. It need scarcely be added that vectors in the 
same direction have the same proportion as the lines which corre- 
spond to them. 

We have then advanced to the following 

LEMMA. All lines drawn in the same direction are, as vectors, 
to be represented by numerical multiples of one and the same 
symbol, to which the ordinary laws of Algebra, so far as their addi- 
tion, subtraction, and numerical multiplication are concerned, may 
be unreservedly applied. 

4. The converse is of course true, that if lines as vectors are 
represented by multiples of the same vector symbol, they are 
parallel. 




8 QUATERNIONS. [CHAP. II. 

It is only necessary to add to what has preceded, that if BC be 
a line not in the same direction with c 

AB, then the vector EG cannot be 
represented by a or by any (arith- 
metical) multiple of a. The vector A 
symbol a must be limited to express transference in a certain 
direction, and cannot, at the same time, express transference in 
any other direction. To express ' vector BC 1 then, another and 
quite independent symbol (3 must be introduced. This symbol, 
being united to a by the signs + and , the laws of algebra will, 
of course, apply to the combination. 

5. If we now join AC, and thus form a triangle ABC, and if 
we denote vector AB by a, BC by ft, AC by y, it is clear that we 
shall be presented with the equation a + ft = y. 

This equation appears at first sight to be a violation of Euclid I. 
20 : " Any two sides of a triangle are together greater than the 
third side". But it is not really so. The anomalous appearance 
arises from the fact that whilst we have extended the meaning of 
the symbol + beyond its arithmetical signification, we have said 
nothing about that of a symbol = . It is clearly necessary that the 
signification of this symbol shall be extended along with that of 
the other. It must now be held to designate, as it does perpetually 
in algebra, ' equivalent to.' This being premised, the equation 
above is f ree.d from its anomalous appearance, and is perfectly con- 
sistent with everything in ordinary geometry. Expressed in words 
it reads thus : ' A transference from A to B followed by a ti-ans- 
ference from B to C is equivalent to a transference from A to C.' 

6. AXIOM. If two vectors have not the same direction, it is 
impossible that the one can neutralize the other. 

This is quite obvious, for when a transference has been effected 
from A to B, it is impossible to conceive that any amount of trans- 
ference whatever along BC can bring the moving point back to A. 

It follows as a consequence of this axiom, that if a, (3 be different 
actual vectors, i. e. finite vectors not in the same direction, and if 



ART. 7.] "VECTOR ADDITION AND SUBTRACTION. 9 

ma. + n{3 = 0, where m and n are numerical quantities ; then must 
m and n = 0. 

Another form of this consequence may be thus stated. If 
[still with the above assumption as to a and /?] ma + n/3 = pa + q(3, 
then must mp, and n q. 

7. We now proceed to exemplify the principles so far as they 
have hitherto been laid down. It is scarcely necessary to remind 
the reader that we are assuming the applicability of all the rules 
of algebra and arithmetic, so far as we are yet in a position to draw 
on them ; and consequently that our demonstrations of certain of 
Euclid's elementary propositions must be accepted subject to this 
assumption. 

To avoid prolixity, we shall very frequently drop the word vector, 
at least in cases where, either from the introduction of a Greek 
letter as its representative, or from obvious considerations, it must 
be clear that the mere line is not meant. The reader will not fail 
to notice that the method of demonstration consists mainly in reach- 
ing the same point by two different routes. (See remark on Ex. 9.) 

EXAMPLES. 

Ex. 1. Tlie, straight lines which join the extremities of equal and 
parallel straight lines towards the same parts are themselves equal 
and parallel. 

Let A E be equal and parallel to CD ; 
to prove that AC is equal and parallel 
to ED. 

Let vector AB be represented by a, 
then (Art. 1) vector CD is also, repre- 
sented by a. 

If now vector CA be represented by (3, vector DB by y, we shall 
have (Art. 5) vector CB = CA + A B = /3 + a, 

and vector CB = CD + DB = a + y ; 
. '. ft + a a + y, 

and (3 = y ; 
so that (3 and y are the same vector symbol; consequently (Art. 1) 




10 QUATERNIONS. [CHAP. TI. 

the lines which they represent are equal and parallel ; i. e. CA is 
equal and parallel to ED. 

Ex. 2. The opposite sides of a parallelogram are equal; and 
the diagonals bisect each other. 

Since AB is parallel to CD, if vector AB be represented by a, 
vector CD will be represented by some numerical multiple of a 
(Art. 3), call it ma.. 

And since CA is parallel to DB; if vector CA be /3, then vector 
DB is nfi ; hence 

vector CB = CA+AB = p + a., 
and = CD + DB = ma + nfi ; 

.-. a + ft = ma + n{3. 

Hence (Art. G) m= 1, n=l, i.e. the opposite sides of the paral- 
lelogram are equal. 

Again, as vectors, AO + OB= AB 

= CD 

= CO + OD ; 

And as AO is a vector along OD, and CO a vector along OB ; 
it follows (Art. 6) that vector AO is vector OD, and vector CO is 
OB; 

O = OD CO = OB. 



Ex. 3. The sides about the. equal angles of equiangular triangles 
are proportionals^ 

Let the triangles ABC, A DE have a common 
angle A, then, because the angles D and B are 
equal, DE is parallel to BC. 

Let vector AD be represented by a, DE by 
/?, then (Art. 3) AB is ma, BC n/3. 

. -. as vectors, AE = AD + DE = a + (3, D 



B 
Now AC is a multiple of AE, call it p(a+{3).- 

. : ma + n[$ = p (a. + (3), 
and m p n (Art. 6). 




EX. 4.] 



VECTOR ADDITION" AND SUBTRACTION. 



11 



But line A B : AD = m, 
line EC : Z>E = n, 
.-. AB : AD :: EC : DE. 

Ex. 4. The bisectors of the sides of a triangle meet in a point 
which trisects each of them. 

Let the sides of the triangle ABC be 
bisected in D, E, F ; and let AD, BE 
meet in G. 

Let vector ED or DC be a, CE or EA (3, F ' 

then, as vectors, 

BA = EC + CA = 2a + 2/? = 2 (a + 0), 




hence (Art. 4) .5^1 is parallel to Z)^, and 
equal to '2DE. 

Again, G+GA= BA 



Now vector _36r is along GE, and vector 6-M along DG. 
.-. (Art. 6) .# 

GA = 
whence the same is true of the lines. 

2 
Lastly, BG = ^BE 



12 QUATERNIONS. [CHAP. II. 

GF=BF-BG 






lience CG is in the same straight line with GF, and equal to IGF. 

Ex. 5. When, instead of D and E being the middle points of 
the sides, they are any points whatever in those sides, it is required 
to find G and the point in which CG produced meets AB. 

BG CA 

Let nr , = m, rr^ n \ a ^ so let vector Z>(7 = a, vector CE ~ (3 ; 
JL/O L/Jli 

.-. BG = ma, CA = n[J. 
Hence BE= BG + CE = 



Let BG = xBE, GA=yDA, 

then BA=BG + GA = x (ma + ft) + y (a -f n/3). 

But BA =ma + n(3, 

. '. (Art. G) xm + y = 



. BG (m-l)n AG (n-l)m 

QYirl 'VIA 11 C\\* - 

* 1 I I lAsm 1. *5* _ _-_ U \J1- . T , , 

BE mn-l AD mn-\. 

Again, let BF=pBA p (ma + nft). 

T) i T> Jjl ~)f*1 i /^ 7^ 

JjUC JjJ. 1 = .DO + O-r 

= ma + a multiple of CG 
= ma + zCG suppose 



= ma + z ' -=- (ma + 8) - 1 

( mn I 

The two values of BF being equated, and Art. 6 applied, 
there results 

??, - 1 m 1 

~ 7/w 1 ' w 1 ' 



EX. 6.] VECTOR ADDITION AND SUBTRACTION. 



13 



whence 



i.e. 



I p n 1 

p nt- 1 ' 
AF AE BD 



or AF . BD.CE=AE.CD. BF. 

Ex. 6. When, instead of as in Ex. 4, where D, E, F are points 
taken within BC, CA, AB at distances equal to half those lines 
respectively, they are points taken in BC, CA, AB produced, at 
the same distances respectively from C, A, and B ; to find tJie inter- 
sections. 

Let the points of intersection be respectively G iy G a , G 3 . 

E 




F^"G' 
Retaining the notation of Ex. 4, we have 



= 3a, CE=3/3; 



and .-. 



and 



= 3a + yD A 



. . 2x = 3 y, 3x = 2y, and x = = : 

7 

.-. line EG 3 =]=EB. 



14 QUATERNIONS. [CHAP. II. 

Similarly line FG l = l - FC, 
\m& DG t =]=DA, 

f 

and from equation (1) EG & = (2a + 3/3). 

But BG a = BA + AG 3 = 2a + 2ft + AG, ; 



2 
hence line J 6r = line DA 



and similarly of the others. 

Ex. 7. :77ie middle points of the lines which join the points of 
bisection of the opposite sides of a quadrilateral coincide, whether 
the four sides of the quadrilateral be in the same plane or not. 

Let A BCD be a quadrilateral ; E, II, G, F the middle points of 
AB, EG, CD, DA X the middle point of EG. 

Let vector AB a, AC = ft, AD = y, 

then AE + G = AD + DG gives / 



= i (* + ft + y), 

which being symmetrical is a, ft, y in the same as the vector to 
the middle point of HF. 

X is called (Art. 14) the mean point of ABCD. 

Ex. 8. The point of bisection of the line which joins the middle 
points of the diagonals of a quadrilateral (plane or not) is the mean 
point. 



EX. 9.] 



VECTOR ADDITION AND SUBTRACTION. 



Let P, Q be the middle points of AC, 
BD, R that of PQ. 

Retaining the notation of the last ex- 
ample we have 




AR=(AP + AQ) 

2i 



Similarly 



i.e. is the same point as X in the last example ; and is therefore 
the mean point of A BCD. 

Ex. 9. AD is drawn bisecting BC in D and is produced to any 
point E ; AB, CE prodded meet in P ; AC, BE in Q ; PQ is 
parallel to BC. 

Let AB = a, AC = ft, 







and AE is a multiple of AD ~ z (a + ft) say. 

Then CP =pC gives xa - ft =p [z (a + j3) - ft], 
.'. (Art. 6) x pz, 1 =pz p ; 

.. p = x+ 1. 

Similarly BQ - qBE gives y(3 a = q {z (a + (3) - a}, 
y = qz, -\=qz-q, 



16 



QUATERNIONS. 



[CHAP. ii. 



i x V i 

and since z = - = we have 

p y 



hence the line PQ is parallel to BG. 

The method pursued in this example leads to the solution of all 
similar problems. It consists, as we have already stated, in reach- 
ing the points P and Q respectively by two different routes, viz. 
through C and through E i or P ; through B and through E for Q 
and comparing the results. 

Cor. 1. PE : EC :: p-\ : 1 :: x : 1 :: AP : AB. 
Cor. 2. AE : AD :: 2z : .1 :: 2x : x+ I 

:: 2(p-l) :p 
:: 1PE : PC, 

.-. AD : DE :: PE + EC : PE-EG. 

Ex. 10. If DEF be drawn cutting the sides of a triangle ; then 
will AD.BF.CE = AE.CF. BD. 



Let BD = a, DA =pa, AE= ft, EG = 



and CF is a multiple of BG. 
Let CF= xBO 




CF=CE + EF 
=-EC+EF 



But 



.'. equating, we have x (1 + p} = yp, x(\+q 
whence x ( 1 +x)pq, 

CF BF AD CE 

1 a _ _ _ 

BG~ BG' BD' AE' 
.-. AD.BF.CE = AE.CF.BD. 



EX. 11.] VECTOR ADDITION AND SUBTRACTION. 17 

Ex. 11. If from any point within a parallelogram, parallels 
be drawn to the sides, the corresponding diagonals of the two 




parallelograms thus formed, and of the original parallelogram 
shall meet in the same point. 

Let PQ, US meet in T; 
join TO, OD. 

Let OA = a, OB = p, OQ=ma, OS=np, 



and 

also FO = TS 

equating, there results 



= TQ-OQ = x{np+(l-m)a}-ma, 



mm- ym ; 



and 



mn 



m 
1 m w' 

mn 



. , 

l-m-n^ ^' I-m-n 

hence (Art. 4) TO, OD are in the same straight line. 

COR. TO : TD :: mn : (l-m)(l-n) :: OSCQ : CRDP. 

Ex. 12. T7te points of bisection of the three diagonals of a com- 
plete quadrilateral are in a straight line. 

1. Q. 2 



18 



QUATERNIONS. 



P, Q, R, the middle points of the 
diagonals of the complete quadrila- 
teral A BCD, are in a straight line. 

Let A = a,AD = (3, 
AE = ma, AF=n'{3; 



D = - ma. and 



gives 
whence 



and 




x (n(3 a) + y (ft ma) = (3 d, 
l, x + my=\, 
ml 



.'. X = 



mn 1 ' 



AP 



=\^-\{ 



1 m (n 1) a + n (m 1) /^ 

2 m/i- 1 



.:AQ-AP = 



AR-AP= 



^ (ma + nft), 



1 



2(wm-l) 



9 limn _L-1 \ I" ' V 

< I llvlb -" J. I 

or vector PR is a multiple of vector PQ, and therefore they are in 
the same straight line. 

COR. Line PQ : PR' :: I : mn 

:: AE.AD : AE . AF 

:: triangle AED : triangle AEF. 

We shall presently exemplify a very elegant method due to 
Sir W. Hamilton of proving three points to be in the same 
straight line. 



ART. 8.] VECTOR ADDITION AND SUBTRACTION. 19 

8. It is often convenient to take a vector of the length of the 
unit, and to express the vector under consideration as a numerical 
multiple of this unit. Of course it is not necessary that the unit 
should have any specified value ; all that is required is that when 
once assumed for any given problem, it must remain unchanged 
throughout the discussion of that problem. 

If the line AB be supposed to be a units in length, and the 
unit vector along AB be designated by a, then will vector AB be 
a (Art. 3). 

Sir William Hamilton has termed the length of the line in 
such cases, the TENSOR of the vector; so that the vector AB is the 
product of the tensor AB and the unit vector along AB. Thus if, 
as in the examples worked under the last article, we designate the 
vector AB by a, we may write a = TaUa, where To. is an abbre- 
viation for ' Tensor of the vector a ; Ua. for ' unit vector along a'. 

EXAMPLES. 

Ex. 1. If tJie vertical angle of a triangle be bisected by a 
straight line which also cuts the base, the segments at the base shall 
have the same ratio that the other sides of the triangle have to one 
another, 

Take unit vectors along AB, AC, which 
call a, /3 respectively : construct a rhombus p. 

APQR on them and draw its diagonal AR. 

Then since the diagonals of a rhombus bi- _ 
sect its angles, it is clear that the vector 

AD which bisects the angle A is a multiple of AR the diagonal 
vector of the rhombus. 

Now AR 




Now vector AB = ca, AC = bfi; using c, b as in ordinary 
geometry for the lengths of AB, AC. 

Hence BD = AD - AB = x (a + /?) - ca, 

and BD = yBC=y(AC-AB] 

- ca). 

22 



20 QUATERNIONS. [CHAP. II. 

Equating, x-c = -yc, x = yb; 



and BD : DC :: y : l-y 

:: c : b 
:: BA : AC. 

COR. If a, ft are unit vectors from A, and if 8 be another 
vector from A such that 8 = x(a + ft); then 8 bisects the angle 
between a and (3. 

Ex. 2. The three bisectors of the angles of a triangle meet in 
a point. 

Let AD, BE bisect A, B and meet in G ; CG bisects C. 

Let units along AH, AC, BC be a, ft, y, then as in the last 

example, 

AG~x(a + ft), BG = y(-a. + y). 

But ay = bft- ca, 

bft-ca 



and CG =AG-AC 

= x( a + ft)-bft, 
also CG^BG-BC 

bft - cd 



=/ a+ 



-ca\ 

/ 



a 

be 

whence x = - - - , 

a + b + c 

and CG 



hence CG bisects the angle C (Cor. Ex. 1). 



AKT. 9.] VECTOR ADDITION AND SUBTRACTION. 21 

9. If a, (3, y are non-parallel vectors in the same plane, it is 
always possible to find numerical values of a, b, c so that aa + b(3 
+ cy shall = 0. 

For a triangle can be constructed whose sides shall be parallel 
respectively to a, (3, y. 

Now if the vectors corresponding to those sides taken in order 
be aa, b{3, cy respectively, we shall have, by going round the 
triangle, 



10. If a, fi, y are three vectors neither parallel nor in the 
same plane, it is impossible to find numerical values of a, b, c, not 
equal to zero, which shall render aa + bft + cy=Q. 

For (Art. 5) aa + b(3 can be represented by a third vector in 
the plane which contains two lines parallel respectively to a, /?. 
Now cy is not in that plane, therefore (Art. 6) their sum cannot 
equal 0. 

It follows that if aa + 6/3 + cy = and a, /?, y are not parallel 
vectors, they are in the same plane. 

11. There is but one way of making the sum of multiples 
of a, (3, y (as in Art. 9) equal to 0. 

Let aa+b/3 + cy = 0, 

and also a + (3 + r = 0. 



By eliminating y we get 

(ar cp)a + (br - cq) ft = ; 
.% (Art. 6) ar cp, br = cg, 
or a : b : c :: p : q : r, 
so that the second equation is simply a multiple of the first. 

12. If a, ft, y are coinitial, coplanar vectors terminating in 
a straight line, then the same values of a, b, c which render 
aa + 6/3 + cy - will also render a -t- b + c = 0. 



22 



QUATERNIONS. 



[CHAP. ii. 




Let vector OA = a, OB = ft, OC = y, ABC 
being a straight line ; then 



But AC is a multiple of AB, 

or y-a=p(p-a), 
i.e. (p l)a pft + y = 0. 
But (;?-l)-p+l=0; 

and as p 1, p, +1 correspond to a, b, c and satisfy the con- 
dition required, the proposition is proved generally (Art. 11). 

13. Conversely, if a, fi, y are coinitial coplanar vectors, and if 
both aa + b(3 + cy = Q and a + b + c - 0, then do a, ft, y terminate 
in a straight line. 

For ay + by + cy = ; 

therefore by subtraction 



i.e. y a is a multiple of y y8, and therefore (Art. 4) in the same 
straight line with it: i.e. AC is in the same straight line with 
BC. (See Tait's Quaternions, 30.) 

EXAMPLES. 

Ex. 1. If two triangles are so situated that the lines which 
join corresponding angles meet in a point, then pairs of correspond- 
ing sides being produced will meet in a straight line. 

ABC, A'B'C' are the triangles; 
the point in which A' A, B'B, C'C 
meet; P, Q, R the points in which 
BC, BC', <fec. meet: PQR is a 
straight line. 

Let OA = a, OB = (3, OC = y, 



and 



BA = a - ft, 
BR = x(a.-p); 
'A' = ma nft, 
B'R = y (ma nfi). 




EX. 2.] VECTOR ADDITION AND SUBTRACTION. 23 

Now BB' = BR- B'R gives 

(n~\.}B = x(a f$) y (ma nf3) ; 
/. n 1 = x + ny, = x my, 

and x = : 



whence OR= OB + BR = B-~ (a- B) 

m n ^ 

_ n (m 1) /3 m (w 1) a 
m . 

Similarly, OP^^^lb^ 



^ _m(p-l)a-p(m-l)y 
p m 

. : (m - n )(p-l)OR+ (v -p)(m- 1) OP 

+ (p-m)(n-l)OQ = Q. 
And also 

(m -n)(p-l) + (n -p) (m - 1) + (p - m) (n-l) = 0, 
whence (Art. 13) P, Q, R are in the same straight line. 

Ex. 2. If a quadrilateral be divided into two quadrilaterals 
by any cutting line, the centres of the three shall lie in a straight line. 

Let Pj(?,$ 3 P 3 be the quadrilateral divided into two by the 




line P S Q 2 . Let the diagonals of P g Q a Q 3 P 3 fflteet in R^ and so of 
the others : R lt R g , R 3 are the centres. 



24 QUATERNIONS. [CHAP. II. 

Produce P 3 P,, Q a Q 1 to meet in 0. Let unit vectors along 
OP, 0$ be denoted by a, ft ; and put 

OP, = w^o, OP, = m s a, OP a = m t a ; 



then OR 3 = OP 1 + P,^ 3 = 7rc,a + x (njl - w^a), 

and OR 3 =OQ^ + Q l R a = 

Equating, we have 

m^-m^x m^/, and 



and Q ^ = OT,W, (n, - n,) a + ra,, (TO, - 

m,ri, - m t w a 
Similarly, 

(m, - 



OR 3 + (mji, - m a n^ m l n l OR^ 

+ (m 3 n 3 w^) m/i a OR S = 0. 
And also 

(mfo - m s n s ) m a n a + (m a n a - m 3 n 3 ) m^ 

+ (m 3 n a -m 1 n l )m a n a = 0, 
whence (Art. 13) R I} R 2 , R a are in the same straight line. 

COR. R t , R 3 , R s will pass through provided the coefficients 
of a and /3 in the three vectors have the same proportion, i.e. 
provided 

I ___ !_ I ___ |_..!_JL. .!_! 
m t m a ' m a m a " w, n a ' n a n a ' 

Ex. 3. If AD, BE, CF be drawn cutting one another at any 
point G within a triangle, tJien FD, DE, EF shall meet the third 
sides of the triangle produced in points which lie in a straight line. 

Also the produced sides of the triangle s/utll be cut harmo- 
nically. 



EX. 3.] VECTOB ADDITION AND SUBTRACTION. 

If, as in Ex. 5, Art. 7, we put 



25 




we get, as in that example, 

AF : BF :: n-l : m-l; 

. .. BF= m ~ l . (ma + ), 

m+n- 2 ^ 



and 



FD=BD- BF= 



DM xFD, compared with 



erives 



- 2) a - 



(m-l)(n-2) (m-l)n 

rg 3 ' V i - J jg _} L___ n I 

m + n 2 m+n 2 



and 



n-2 



ft- 



n-l 



Again, FE=FA+AE = - {ma - (m - 2) $} 
m + n-2 ( 



26 QUATERNIONS. [CHAP. II. 

And EL = xFE, compared with 



m 
gives y 



m(m-l) 

a. 



Thirdly, i)jV= xZ>^ = a; (a + /3), compared with 

= BN- BD = y (ma + nj3) -(m-l) a, 
m-I 



gives y 



m n 



and EN = - (ma + n(3). 

m-n^ 

Now (m -l)(n- 2) BM + (m - n) BN 



Also (m-l)(n- 2) + (m- n)- (m- 2) (n-I) = Q 
therefore BM, BN, BL are in a straight line (Art. 13). 

Further, CL = ^ CD, 

m 2 



m-2 

.-. CL : CD :: BL : BD, 
and BL is cut harmonically. 

Ex. 4. The point of intersection of bisectors of tJie sides of a 
triangle from the opposite angles, the point of intersection of per- 
pendiculars on tJie sides from the opposite angles, and the point of 
intersection of perpendiculars on tJie sides from their middle points, 
lie in a straight line which is trisected by the first of these points. 

1. Let unit vector CJ3 = a, unit vector CA = (3, 
then, Ex. 4, Art. 7, CG = \ (aa + bft). 



ART. 14.] VECTOR ADDITION AND SUBTRACTION. 27 

2. Let AH, BK perpendiculars on the A 

sides intersect in 0, /\\ ft 

then HA = bft-bacosC, / JAY 

= b(ft-a cos C), 



Now CO = CA +AO, and also = CB + SO, gives 

6)3 + yb (ft - aa cos C) = aa + xa(a ft cos (7), 
6 cos C a 




and CO = -. ~ {(6 a cos (7) a + (a - b cos C) ft}. 

sin 2 (7 lv 

3. Let perpendiculars from I) and E (-Ex. 4, Art. 7) meet 
in X, 

then DX is a multiple of HA. 

. : CX= CD + DX = CE + EX gives 

^ aa + v (ft - a cos C) = ^ bft + z (a. - ft cos C), 

2t a 

b a cos C 



2 sin 2 C 

(a b cos C) a + (b a cos (7) ft 
2 sin 2 C ~~ 1 



and C X = 



and also 2 + 1-3 = 0, 

.. X, 0, G are in a straight line. 
Also CO-CG=2 (CG - CX), 

or vector GO = 2 vector XG, 



and G trisects XO. 

14. The vector to the mean point of any polygon is the mean 
of the vectors to the angles of the polygon. 



28 QUATERNIONS. [CHAP. II. 

1. Let be any point ; then in the figure of Ex. 4, Art. 7, 
we have, calling OA, a, OB, (3 and 00, y, 

OG=a+AG=ft+G=y+CG 



1 

9 

because AG + BG + CG =-- 1 (AD + BE + CF) 

o 

= I {(AB + AC) + (BA + C) + (CA + CB)} 
= 0. 

2. If OA, OB, OC, OD be a, ft, y, 8, in the figure of Ex. 7, 
Art. 7, we have 

= OH + HX=OH+l (OF- OH) 



3. In the more general case we may define the mean point in 
a manner analogous to that adopted in mechanics to define the 
centre of inertia of equal masses placed at the angular points of 
the figure. Thus, if we take any rectangular axes OX, OY, and 
designate by a, ft unit vectors parallel to these axes; and by p,, 
p 4 , &c. the vectors to the different points; and if we write x^ y,; 
x ii y> & c - f r the Cartesian co-ordinates of the different points 
referred to those axes ; and define the mean point as the centre of 
inertia of equal masses placed at the angular points; the Cartesian 
co-ordinates of that point will be 

a?, + ,+ ... _ 

~ ~ 



and its vector p = xa + yft. 



EX. 1.] VECTOR ADDITION AND SUBTRACTION. 29 

Now p l = xp + yfi, p a = x,a + y, &c. 



g ^ , , ... 
m " 



= /> 

COE. 1. ( Pl - P ) + (p 2 -p) + (p 3 
i. e. the sum of the vectors of all the points, drawn from the mean 
point, = 0. 

The extension of the same theorem to three dimensions is 
obvious. 

COR. 2. If we have another system of n points whose vectors 
are cr l , o- , &c. then the vector to the mean point is 



n 

If now T be the mean point of the whole system, we have 
T== Pi + P + +<r 1 + <r,+ ... 



or (m + n) r mp rw = 0, 

hence (13) T, p, cr terminate in a right line; or the general mean 
point is situated on the right line which connects the two partial 
mean points. 

ADDITIONAL EXAMPLES TO CHAP. II. 

1. If P, Q, B, S be points taken in the sides AB, EG, CD, 
DA of a parallelogram, so that AP : AB :: BQ : BC, &c., PQRS 
will form a parallelogram. 

2. If the points be taken so that AP = CR, BQ = DS, the 

same is true. 

3. The mean point of PQRS is in both cases the same as that 
of ABCD. 



30 QUATERNIONS. [CHAP. II. 

4. If FQ'R'S' be another parallelogram described as in Ex. 1, 
the intersections of PQ, P'Q', <fec. shall be in the angular points of 
a parallelogram EFGH constructed from PQRS as P'Q'R'S' is 
constructed from ABGD. 

5. The quadrilateral formed by bisecting the sides of a 
quadrilateral and joining the successive points of bisection is a 
parallelogram, with the same mean point. 

6. If the same be true of any other equable division such as 
trisection, the original quadrilateral is a parallelogram. 

7. If any line pass throvigh the mean point of a number of 
points, the sum of the perpendiculars on this line from the 
different points, measured in the same direction, is zero. 

8. From a point E in the common base AB of the two 
triangles ABC, ABD, straight lines are drawn parallel to AC, AD, 
meeting BC, BD at F, G ; shew that FG is parallel to CD. 

9. From any point in the base of a triangle, straight lines are 
drawn parallel to the sides: shew that the intersections of the 
diagonals of every parallelogram so formed lie in a straight line. 

10. If the sides of a triangle be produced, the bisectors of the 
external angles meet the opposite sides in three points which lie 
in a straight line. 

11. If straight lines bisect the interior and exterior angles 
at A of the triangle ABC in D and E respectively; prove that BD, 
BC, BE form an harmonica! progression. 

12. The diagonals of a parallelepiped bisect one another. 

13. The mean point of a tetrahedron is the mean point 
of the tetrahedron formed by joining the mean points of the 
triangular faces ; and also those of the edges. 

14. If the figure of Ex. 11, Art. 7, be that of a gauche quadri- 
lateral (a term employed by Chasles to signify that the triangles 



EX. 15.] VECTOR ADDITION AND SUBTRACTION. 31 

AOD, BOD are not in the same plane), the lines QP, DO, RS will 
meet in a point, provided 

AP OS . AQ DR 



15. If through any point within the triangle ABC, three 
straight lines MN, PQ, RS be drawn respectively parallel to the 
sides A B, AC, BC ; then will 

MN P RS 



_ 
AB 1U JfU~~' 

1C. A BCD is a parallelogram; E, the point of bisection of 
AB ; prove that AC, DE being joined will trisect each other. 

17. ABCD is a parallelogram ; PQ any line parallel to CD ; 
PD, QC meet in S, PA, QB in R prove that AD is parallel to 

ss. 



CHAPTER III. 

VECTOR MULTIPLICATION AND DIVISION. 

15. WE trust we have made the reader understand by what we 
stated in our Introductory Chapter, that, whilst we retain for 
'multiplication' all its old properties, so far as it relates to ordi- 
nary algebraical quantities, we are at liberty to attach to it any 
signification we please when we speak of the multiplication of a 
vector by or into another vector. Of course the interpretation of 
our results will depend on the definition, and may in some points 
differ from the interpretation of the results of multiplication of 
numerical quantities. 

It is necessary to start with one limitation. Whereas in 
Algebra we are accustomed to use at random the phrases ' multiply 
by' and 'multiply into' as tantamount to the same thing, it is 
now impossible to do so. We must select one to the exclusion of 
the other. The phrase selected is 'multiply into'; thus we shall 
understand that the first written symbol in a sequence is the 
operator on that which follows : in other words that a/2 shall read 
'a into /?', and denote a operating on /?. 

16. As in the Cartesian Geometry, so v 
here we indicate the position of a point in 

space by its relation to three axes, mutually 
at right angles, which we designate the axes 
of x, y, and * respectively. For graphic 
representation the axes of x and y are 
drawn in the plane of the paper whilst that 
of z being perpendicular to that plane is 
drawn in perspective only. As in ordinary 



ART. 17.] VECTOR MULTIPLICATION AND DIVISION. 33 

geometry we assume that when vectors measured forwards are 
represented by positive symbols, vectors measured backwards will 
be represented by the corresponding negative symbols. In. the 
figure before us, the positive directions are forwards, upwards 
and outwards; the corresponding negative directions, backwards, 
downwards and inwards. 

With respect to vector rotation we assume that, looked at in. 
perspective in the figure before us, it is negative when in the 
direction of the motion of the hands of a watch, positive when in 
the contrary direction. In other words, we assume, as is done in 
modern works on Dynamics, that rotation is positive when it 
takes place from y to z, z to x, x to y : negative when it takes 
place in the contrary directions (see Tait, Art. 65). 

Unit vectors at riglvt angles to each otJier. 

17. DEFINITION. If i, j, k be unit vectors along Ox, Oy, Oz 
respectively, the result of the multiplication of i into j or ij is 
defined to be the turning of j through a right angle in the plane 
perpendicular to i and in the positive direction ; in other words, 
the operation of i on j turns it round so as to make it coincide 
with k ; and therefore briefly ij = k. 

To be consistent it is requisite to admit that if i instead of 
operating on^' had operated on any other unit vector perpendicular 
to i in the plane of yz, it would have turned it through a right angle 
in the same direction, so that ik can be nothing else than j. 
Extending to other unit vectors the definition which we have 
illustrated by referring to i, it is evident that j operating on k 
must bring it round to i, orjk i. 

Again, always remembering that the positive directions of 
rotation are y to z, z to x, x to y, we must have ki =j. 

18. As we have stated, we retain in connection with this 
definition the old laws of numerical multiplication, whenever 
.numerical quantities are mixed up with vector operations; thus 
2i . 3j = Gij. Further, there can be no reason whatever, but the 
contrary, why the laws of addition and subtraction should undergo 

T. Q. 3 



34 QUATERNIONS. [CHAP. III. 

any modification when the operations are subject to this new 
definition ; we must clearly have 



Finally, as we are to regard the operations of this new de- 
finition as operations of multiplication magnitude and motion 
of rotation being united in one vector symbol as multiplier, 
ju*t as magnitude and motion of translation were united in 
one vector symbol in the last chapter we are bound to retain 
all the laws of algebraic multiplication so far as they do not 
give results inconsistent with each other. In no other way can 
the conclusions be made to compare with those deduced from 
the corresponding operations in the previous science. Thus we 
retain what Sir William Hamilton terms the associative law of 
multiplication : the law which assumes that it is indifferent in 
what way operations are grouped, provided the order be not 
changed ; the law which makes it indifferent whether we consider 
a be to be a x be or ab x c. This law is assumed to be applicable to 
multiplication in its new aspect (for example that ijk ~ ij . &), and 
bding assumed it limits the science to certain boundaries, and, 
along with other assumed laws, furnishes the key to the interpreta- 
tion of results. 

The law is by no means a necessary law. Some new forms of 
the science may possibly modify it hereafter. In the meantime 
the assumption of the law fixes the limits of the science. 

The commutative law of multiplication under which order may 
be deranged, which is assumed as the groundwork of common 
algebra (we say assumed advisedly) is now no longer tenable. And 
this being the case it is found that the science of Quaternions 
breaks down one of the barriers imposed by this law and expands 
itself into a new field. 

ij is not equal toji, ib is clearly impossible it should be. 

A simple inspection of the figure, and a moment's consideration 
of the definition, will make this plain. The definition imposes on i 
as an operator on^' the duty of turning^' through a right angle as 
if by a left-handed turn with a cork-screw handle, thus throwing 
j up from the plane xy; when, on the other hand,J is the operator 



ART. 19.] VECTOR MULTIPLICATION AND DIVISION. 35 

and i the vector operated on, a similar left-handed turn will bring 
i down from the plane of xy. In fact ij = Jc, ji = k, and so 

y = -ft- 

19. We go on to obtain one or two results of the application 
of the associative law. 

1. Since ij = k, we have i . ij = ik = j. 
Now by the law in question, 



or i = l. 

Our first result is that the square of the unit vector along Ox 
is 1 ; and as Ox may have any direction whatever, we have, gene- 
rally, the square of a unit vector = 1. In other words, the 
repetition of the operation of turning through a right angle reverses 
a vector. 

2. Again, ijk = i .jk = i . i = i 2 = 1. 
Similarly it may be proved that 

jki = kij = -l, 

or no change is produced in the product so long as direct cyclical 
order is maintained. 

3. But ikj=i . kj = i . i tf = + 1 ; 

.-. ijk^-ikj, 

or a derangement of cyclical order changes the sign of the product. 
This last conclusion is also manifest from Art. 18. 

Vectors generally not at right angles to each other. 

20. We have already (Art. 8) laid down the principle of 
separation of the vector into the product of tensor and unit 
vector ; and we apply this to multiplication by the considerations 
given in Art. 18, from which it follows at once that if a be a 
vector along Ox containing a units, /? a vector along Oy con- 
taining b units, 

a = ai, ft = bj, and a/? = abij. 

32 

t 



36 QUATERNIONS. [CHAP. III. 

In the same way 

a 2 = ai . ai = a*i 2 = a 2 , 

or the square of a vector is the square of the corresponding line 
with the negative sign. 

Seeing therefore the facility with which we can introduce 
tensors whenever wanted, we may direct our principal attention, 
as far as multiplication is concerned, to unit vectors. 

21. We proceed then next to find the product a/3, when a 
and /3 are vectors not at right angles to one another. 
1. Let a, ft be unit vectors. 
Let OA - a, OB = ft. 

Take OC = y, a unit vector perpen- 
dicular to OB and in the plane BOA. 
Take also DO or DO produced- e, a unit 



vector perpendicular to the plane BOA. 




Draw AM, AN perpendicular to OB, 
OC, and let the angle BO A = ; then 

vector OA = OM + MA = OM+ ON (Art. 1) 
= part of OB + part of OC (Art. 3). 

Now it is evident that OM as a line is that part of OB which 
is represented by the multiplier cos 6, or OM = OB cos 9, and 
similarly that ON=OCs\nO: consequently (Art. 3) the same 
applies to them as vectors ; i. e. 

vector OM=(3cosO, vector ON=y sin 6} 

.'. a = (3 cos 6 + y sin 6, 

and a/3 = (/3 cos + y sin 0) /3 

= /3 2 cos + y/3 sin 0. 
But /3 2 = -l (19. 1), 

y/3 = e (17); 

[Observe that y, (3 and c of the present Article correspond 
toj, i and -k of Art. 17.] 

.'. aft cos 0+ esintf. 



ART. 22.] VECTOR MULTIPLICATION AND DIVISION. 37 

2. If a, /3 are not unit vectors, but contain To. and Tft units 
respectively, we have at once, by the principle laid down in 
Art. 20, 

a/2 = TaTfi (- cos + e sin 0). 

3. It thus appears that the product of two vectors a, /3 not 
at right angles to each other consists of two distinct parts, a 
numerical quantity and a vector perpendicular to the plane of 
a, /?. The former of these Sir William Hamilton terms the SCALAR 
part, the latter the VECTOR part. We may now write 

a/3 = Saj3 + Fa/8, 
where S is read scalar, F vector : and we find 



7afi = TaTfi sin 0. 

4. The coefficient of e in Fa/3 is the area of the parallelogram 
whose sides are equal and parallel to the lines of which a, /? are 
the vectors. 

22. To obtain /3a we have, a and (3 being unit vectors, 
a = /? cos 6 + y sin 6 ; 



- - cos - e sin (Art. 19. 1 and 18) ; 
therefore generally 

(3a = TaTft (- cos - e sin 0). 

It is scarcely necessary to remark that whilst y operating on 
ft turns it inwards from OB to DO produced, /? operating on y 
turns it outwards from 00 to OD, causing it to become - e. 

We have therefore 

1. Sap = S(3a. 

2. 7ap = -V{3a. 

3. ap + fia = 2Sa/3. 

4. ap-p a =2Vap. 

/1C 



38 QUATERNIOXS. [CHAP. III. 

5. a + P=a + Pa + P 



6. (a-pY = a 3 -2Sap+p 2 . 

7. If a, (3 are at right angles to each other, Saj3 = 0, and 

conversely. 

8. Vap is a vector in the direction perpendicular to the 

plane which passes through a, /?. 

9. a*/3* = a/3 . Pa because /3* is a scalar ; 

af3 - Va0) 



Note, a 2 ft 2 must not be confounded with (aft) 3 . 

23. Before proceeding further it is desirable we should work 
out a few simple Examples. 

Ex. 1. To express the cosine of an angle of a triangle in terms 
of the sides. 

Let ABC be a triangle ; and retaining the usual notation of 
Trigonometry, let 

CB=a, CA=/3; 
then (vector A) s = (a - /?)' 

= a s -2Sa.p + p* (22. 6), 

or, changing all the signs to pass from vectors to lines (20) and 
applying 21. 3, 



Ex. 2. To express the relations between the sides and opposite 
angles of a triangle. 

Let CB = a, CA = p, JBA = y. 

Then CB + BA = CA gives 



. . a* = a (P y) a/3 ay. 
Take the vectors of each side. 



AET. 23.] VECTOR MULTIPLICATION AND DIVISION. 30 

Xow Fa* = 0, for a 2 = - a 3 has no vector part, 



i e. (21. 3) abe sin C = ace sin JS, 

or b sin C c sin B ; 
Le. b : c :: sinJ5 .: sin (7. 

Ex. 3. TAe sum of the squares of the diagonals of a paral- 
lelogram is equal to t/te sum of the squares of the sides. 
Retaining the notation and figure of Ex. 1, Art. 7, 



.'. CB* + DA 3 = 2a 2 + 2^, 

and, changing all the signs, we get (20) for the corresponding 
lines, 



Ex. 4. Parallelograms upon the same base and between t/te 
same parallels are equal. 

It is necessary to remind the reader of what we have already 
stated, that examples such as this are given for illustration only. 
We assume that the area of the parallelogram is the product of 
two adjacent sides and the sine of the contained angle. 

Adopting the figure of Euclid I. 35 and writing TVfia. as the 
tensor multiplier of FySa so as to drop the vector e on both sides; 
we have, calling LA, a ; BC, ft ; 

BE=BA+AE 



.e. a 

remembering that je/3 2 has no vector part, 
Hence T.Vfta^T (BC . BE), 

i. e. BC . BA sin ABC = BC.BE sin EEC (21. 3), 
which proves the proposition. 



40 QUATERNIONS. [CHAP. III. 

Ex. 5. On the sides AS, AC of a triangle are constructed any two 
parallelograms ABDE, ACFG : the sides DE, FG are produced to 
meet in II. Prove that the sum of the areas of the parallelograms 
ABDE, ACFG is equal to the area of the parallelogram whose 
adjacent sides are respectively equal and parallel to BC and AH. 

Let BA = a, AE=$, AC = y, GA=S, 

then AH = (3 + xk, and AH= $yy; 

.: VaAH = Vap and VyAH=-VyS 

= V8y(22. 2), 

hence F (a + y) AH= Vap + FSy, 

i. e. (21. 4), the parallelogram whose sides are parallel and equal to 
BC, AH, equals the two parallelograms whose sides are parallel 
and equal to BA^ AE $ GA, A C respectively. 

[The reader is requested to notice that the order GA, AC is the 
same as the order BA, AE, and BA, All : so that the vector e 
is common to all.] 

Ex. 6. If be any point whatever either in the plane of the 
triangle ABC or out of that plane, the squares of the sides of the 
triangle fall short of three times the squares of the distances of the 
angular points from 0, by the square of three times the distance of 
the mean point from 0. 

Let OA = a, OB = p, OC = y, 

then (Art. 14), OG = \ (a + p + y), 

o 

or a* + (l 2 + y a + 2S(ap + l3 7 + ya) = 30G*. 

Now AB=p-a, C = y-p, CA=a-y, 

. '. AB 2 + BC 1 + CA 2 = 2 (a 2 + yS 2 + y 2 ) - 2S (a + /3y + ya) 



and the lines 

AB* + BC 2 + CA 3 = 3 (OA* + OB 2 + OC 2 ) - (30G)*. 

Ex. 7. The sum of the squares of the distances of any point 
from the angular points of the triangle exceeds the sum of the 



ART. 23.] VECTOR MULTIPLICATION AND DIVISION. 41 

squares of its distances from the middle points of the sides by the 
sum of the squares of half the sides. 

Retaining the notation of the last example, and the figure of 
Ex. 4, Art. 7, 

OZ) = l(/3 + y) > OE= l -(y + a\ 0^=1 ( + ); 
.'. 4 (OD* + OE 2 + OF 9 ) = 2 (a 2 + /3 2 + y 2 ) + 2S (a/? + Py + ya) 



= 4 (a 2 + ft 2 + y 2 ) - (AB 2 + BC 2 + CA 2 ) ; 

A 7? 2 4- T?^ 8 -i- ^y4* 
.-. as lines OD 2 + OE 2 + OF 1 + + oyi = ^^ + O^ 2 + (9C 12 . 



Ex. 8. IVie squares of the sides of any quadrilateral exceed the 
squares of the diagonals by four times the square of the line which 
joins the middle points of the diagonals. 

Retaining the figure and notation of Ex. 8, Art. 7, we have 
squares of sides as vectors 



and squares of diagonals 



therefore the former sum exceeds the latter by 



Tlierefore as lines the same is true. 

Note. The points A, B, C, D need not be in one plane. 



QUATERNIONS. 



[CHAP. in. 



Ex. 9. Four times the squares of the distances of any point 
whatever from the angular points of a quadrilateral are equal to thi 
sum of the squares of the sides, the squares of the diagonals and the 
square of four times the distance of the point from the mean point 
of t/ie figure. 

With the notation of Art. 14, and the figure of Ex. 7, Art. 7> 
we have 

squares of the sides + squares of the diagonals 



- 3 (a 2 + ft 2 + y 2 + o 2 ) - 2S (a/3 + ay + aS + /3y + 5 + yS). 
Now (Art. 14) (a + /? + y + S) 2 = (WX) 2 j 

. '. (4CLr) 8 + squares of sides -f squares of diagonals 
= 4 (OA* + OB 2 + OC 2 + OD*). 

Ex. 10. The lines which join the mean points of three equila- 
teral triangles described outwards on the three sides of any triangle 
form an equilateral triangle whose mean point is the same as that of 
the given triangle. 

Let P, Q, R be the mean points of the equilateral triangles on 
BG, CA, AB; PD= a, DC - (3, CE = y , EQ = 8 ; and let the sides 
of the triangle ABC be 2a, 26, 2c. 




ft 2 -f 



ART. 23.] VECTOR MULTIPLICATION AND DIVISION. 43 

Changing all the signs and observing that 

2 
/Sa(3 0, Say = -- -p; ab sin C, &c. 

V 

we have (writing the results in the same order), 

Q 2 = ~ + a 2 + b s + ^ + 
o o 

22 2 

+ . ab sin C + -= ab cos C 2ab cos C + - . ab sin C + 

V'" " v " 

4 4 

= K ( 2 + 6 2 - ab cos C) + 7^06 sin C 
> vi 

= | (a 2 + 6 2 + c 2 ) + ~ area of ^ BC, 
o Jo 

which being symmetrical in a, b, c proves that PQR is equilateral. 
Again, G being the mean point of ABC, 



i T 7J/-V2 a 2 a 2 46 2 4 _, 4 . 

and line PG =-^- + -& + - - + ^ 7^ a6 sin C - r a5 cos C 

o y y o ,y/o y 

- (a 2 + 6 2 + c 2 ) + area 



and 6f is the mean point of the equilateral triangle PQR. 

Ex. 11. In any quadrilateral prism, tlie sum 
of the squares of the edges exceeds the sum of the F , 
squares of the diagonals by eight times the square 
of the straight line which joins the points of inter- 
section of t/ie two pairs of diagmials. 



sum of squares of edges = 

2 {a 2 + 2 + (y- a) 2 + (y - /5) 2 + 28 2 } 

= 2 (2a 2 + 2 ft 3 + 2y 2 + 28 2 - 2Say 







44 QUATERNIONS. [CHAP. III. 

sum of squares of diagonals 

= (S + 7 ) 2 + (S- 7 ) 2 +(S + a-y8) 2 + (S + /3-a) 2 
= 2 {a 2 + ft 3 + / + 2S 2 - 2Sa/3}. 

Also 10Gf = l(8 + y ) 

= vector to the point of bisection of 

CD, and therefore to the point of intersection of OG, CD, 
and vector from to the point of bisection of AF, as also to that 
of BE, and therefore to the intersection of A F, BE 



hence vector which joins the points of intersection of diagonals 



eight times the square of this vector 

= 2 (a 2 + P 2 + / + 2Sap - 2Say - 2Spy), 

which, added to the sum of the squares of the diagonals, makes up 
the sum of the squares of the edges. 

24. DEFINITION. We define the quotient or fraction , where 
a and p are unit vectors, to be such that when it operates on a it 
produces p or . a = /?. This form of the definition enables us to 
strike out a by a dash made in the direction of ordinary writing, 

thus . a = p. is therefore that multiplier which, operating 
a a. 

on a, or on p cos + y sin (21), produces p. 

Now cos + e sin operating on p cos + y sin produces 

P cos 8 + (y + e/3) sin cos + ey sin 2 0. 
But a glance at the figure (Art. 21) will shew that 



and 



ART. 25.] VECTOR MULTIPLICATION AND DIVISION. 45 

.-. cos 6 + e sin operating on /? cos + y sin produces /3 ; 

hence = cos + c sin 0. 

a 

It may be worth while to exhibit another demonstration of 
this proposition : thus 

- . a/2 = (3 . ft (by the associative law) = - 1 . (19 . 1). 

i.e. (21 . 1) . (-cos0 + esin0) = -l. 

Now (cos 6 + e sin 6) ( cos + e sin 6) 

= - cos 2 sin 2 

i . 

~ A > 

.*.= cos + e sin 0. 



a 



COR. = -a(by 22). 






25. 1. DEFINITION. Still retaining a, (3 as unit vectors, since 
operating on a causes it to become /3, it may be defined as a 

VERSOR acting as if its axis were along OD (Fig. Art. 21). By 
comparing the result of that article with the definitions of Art. 

17, it is clear that or cos + c sin is an operator of the same 

character as k or e (as we have now called the corresponding 
unit vector) ; with this difference only, that whereas k or c as an 
operator would turn a through a right angle, cos + e sin 9 turns it, 
in the same direction, only through the angle : cos 6 + e sin 6 is 
then the versor through the angle 0. 

2. If a, ft are not unit vectors, the considerations already 
advanced render it evident that 



TB 
Now j~- is itself of the nature of a tensor, for it is a numerical 

J.O. 

quantity, hence - is the product of a tensor and a versor. 




46 QUATERNIONS. [CHAP. III. 

26. By comparing the last Article with Art. 22 it appears 
that generally the product or quotient of two vectors may be 
expressed as the product of a tensor and a versor. This product 
Sir W. Hamilton names a QUATERNION. 

COR. It is evident that a quaternion is also the sum of a 
scalar and a vector. 

27. (1) If a > A 7 are un it vectors in the same plane, c a 
unit vector perpendicular to that plane ; we 

have seen that - operating on a turns it 
round about e as an axis to bring it into the 
position /?. If now - be a second operator 

about the same axis in the same direction 

acting on (3, it will bring it into the position y. But it is evident 

that - acting on a would at once have brought it into the position 
a 

y. This is equivalent to the fact that ^ . = - ; or in another 

pa a 

form (Art. 24) that 

(cos < + c sin </>) (cos 9 + c sin 9} = cos (9 + </>) + c sin (9 + </>). 

Prom this it is evident thnt the results of Demoivre's Theorem 
apply to the form cos 9 + c sin 9. 

Further, it is evident that since cos 9 + e sin 9 operating with c 
as its axis, turns a vector through the angle 9, whilst e itself acting 
in the same direction turns it through a right angle, cos 9 + c sin 9 
is part of the operation designated by e, viz. that part which bears 
to the whole the proportion that 9 bears to a right angle. 

(2) Remembering then that the operations are of the nature 
of multiplication, it becomes evident that cos 9 + c sin 9 as an 

~ 29 

operator may be abbreviated by or e w . 
And since 
(cos 9 + e sin 9} (cos < + e sin <) = cos (9 + <) + e sin (0 + <), 



ART. 28.] VECTOR MULTIPLICATION AND DIVISION. 47 

we shall have 



or the law of indices is applicable to this operator. 

(3) Now we have already seen (19. 1) that c 2 = 1 ; 

.'. 4 = + l. 

Conversely, if c" = e, n must be an odd number; if e" = -l, 
n must be an odd multiple of 2 ; and if c" = + 1, n must be an even 
multiple of 2. 

(4) "When a, (3 are not units, the introduction of the corre- 
sponding tensor can be at once effected. 

We conclude that a quaternion may be expressed as the power 
of a vector, to which the algebraic definition of an index is 
applicable. 

28. Reciprocals of quaternions unit vectors. 

1. Since a.a = o s = 1, 

and -.a=l (Def. Art. 24) 

= a . a ', 

.: - = a, or a" 1 =-a: 
a 

or the reciprocal of a unit vector is a unit vector in the opposite 
direction. 

2. Again, a.- = a(-a) = l=-.a; 

a a 

or a vector is commutative with its reciprocal. 

3. If q be a versor ( say cos + e sin 0, or - J , 

- . q = 1 (Def. extended). 

Now = q ; 

a. 

.'.ft- qa, by operating on a. 



48 QUATERNIONS. [CHAP. III. 

a 1 

Also -^ = - , 

a = - /?, by operating on {3, 

and /3 = g-a = q . - (3 ; 

1 1 

. '. q . - = 1 = - . q, 
q q ' 

or q and - are commutative. 

This is perhaps better demonstrated by observing that 

~ ' a~ a~~ > 

a p p 

or that if = cos + e sin 6, 

a 

then must -= = cos Q e sin 6 ; 

factors which are from their very nature commutative. 
Asa verification, we have 

.75= (cos 6 + sin 6) (cos - e sin 6) 
a }3 

= (cos e) 2 - c 2 (sin ey 

because e 2 = - 1 (28. 1). 

When the versors are not units the tensors can be introduced 
as mere multipliers without affecting the versor conclusions. 

29. We present one or two examples of quaternion division. 

Ex. 1. To express sin (0 + <) and cos (0 + <) in terms of sines 
and cosines of 6 and <. 

a, /?, y being unit vectors in the same plane (Fig. Art. 27), we 
have 

- = cos + c sin 6, 
a. 



ART. 29.] VECTOR MULTIPLICATION AND DIVISION. 49 

y . 

jr = cos <f> + e sin <f>, 

2 = cos (0+ <f>) + sin (&+ <). 

But ?=!.; 

a (3 a 

. . cos (5 + <) + 6 sin (# + <) = (cos + e sin 0) (cos < + e sin <) ; 
whence multiplying out and equating, we have 

sin (6 + <) = sin 6 cos < + cos 6 sin <, 
cos (6 + <) = cos cos < sin 6 sin <. 

COR. If the action of the versors be in opposite directions, 
y3 lying beyond y, \ve have (Art. 28) 

- = cos (0 -<}>)- sin (0 - <). 
But - = cos d> + e sin d>. 

y 

-~ = cos - e sin ; 
p* 

a a B . 
' - -7i - Rives 

y 3 y * 

cos (0 - <) - sin (0 - <) = (cos - sin 6) (cos + e sin <), 
whence sin (0 <) = sin cos ^> cos sin <f> , 

cos ($-<{)) = cos cos /> + sin 5 sin <. 

Ex. 2. To ^%^ tlte cosine of the angle of a spherical triangle 
in terms of the sides. 

Let a, (3, y be unit vectors OA, OS, OC not in the same 
plane, then 




. 

i.e. taking the scalar of each side, 

a fvP v a \ - 

cos a ~ cos c cos 6 + o ' . ( V - . V - J B 

T. Q. 

. T. Co 



50 QUATERNIONS. [CHAP. III. 

Now /SV V is sin c sin b x cosine of the angle between 
a y 

perpendiculars to the planes AD, AC, and is therefore 

sin b sin c cos A ; 
/. cos a = cose cos b + sin c sin b cos A. 

The reader will observe that in accordance with the results of 
Art. 21, the sign of the term involving cos .4 is +, seeing that it is 
in fact cosine (supplement of A). 

Ex. 3. The angles of a triangle are together equal to two right 
angles. 

What we shall prove in fact is that the exterior angles formed 
by producing the sides in the same direction are equal to four 
right angles. 

Let unit vectors along BC, CA, AB be a, /?, y ; and let the 
exterior angles formed by producing BC, CA, AB be 0, </>, i/^; 
then 

29 
e"a = /3(27. 1), 



24> 29 2 

.'. e"" . t.* a = c' r 

2<ft 2j 29 2j 

and e* 1 . e* . "' a = ' r 

2^ 2^1 29 

so that " . e 17 . e 71 " = 1, 



= 1 (27. 2). 
2 
Hence (27. 3), - (0 + < + iff) is an even multiple of 2. The 

first value is 4 ; 

or the exterior angles of a triangle are equal to four right angles. 



ART. 29.] VECTOR MULTIPLICATION AND DIVISION. 51 

It will be seen that the demonstration here given is of the 

nature of that given by Prof. Thomson in the Notes to his Euclid. 

i 

[More directly 






From these 



or A + + = ir.] 

Ex. 4. In the figure of Euclid i. 47 the three lines AL, BK, 
GF meet in a point. 

Let BC = a, CA = /?, AB = y; the sides being as usual denoted 
by a, b, c. 

Let i be the vector which turns another negatively through a 
right angle in the plane of the paper, so that 



If BK, AL meet in 0, 

BO 
and BO 



x (a + iff) = y + yia, 
xSa (a. + ifi) = - Say, 

Say etc cos B 

' - 



- - - 

Sa (a + if}) a 2 + ab sin 
c> 



and xSa/3 = ySia/3 

b be 



42 



52 QUATERNIONS. [CHAP. III. 

which being symmetrical in b and c shews that CF, AL intersect 
in the same point in which BK, AL intersect. 

BO _ c 3 

C*OR. feince ,, fi 

BK 

CO 

we have 



also 



CF a*+bc' 

AO bo 

BD ~ a" + be ' 

AO^ B0_ CO _ c a + b* + be 
'' BD + ~BK + CF~ a* + bc 

Ex. 5. If A BCD be a quadrilateral inscribed in a circle ; 



Let unit vectors along AB, BC, CD, DA be a', ft', y, X ; and 
let the exterior angles at B and D be 6 and < respectively ; then 

a'py = (- cos 6 + e sin 6) y' (21. 1) 
= (cos (ft + e sin 
= 8' (25. 1); 
therefore, introducing the tensors, 



Conjugate Quaternions. 

30. If we designate by y the expression cos + e sin 0, we 
have seen that it may be regarded as a versor through an angle 
in a certain direction. Now if we write in place of 6 in this 
expression it assumes the form cos c sin 0, which must on 
the same hypotheses be regarded a versor through the angle 6 in 
the contrary direction. 

When the quaternion is completed by the introduction of a 
tensor Tq, if we retain the same tensor to both forms of the 



ART. 30.] VECTOR MULTIPLICATION AND DIVISION. 53 

versor, we have Sir "W. Hamilton's conjugate quaternion defined 
thus : The conjugate of a quaternion q, written Kq, has the same 
tensor, plane and angle as q has, only the angle is taken in the 
reverse way. 

The analogy between q and Kq is precisely the same as that 
which exists between the two forms 



R (cos </> + - 1 sin <) and R (cos < J - 1 sin <) ; 

and as the product of the latter form is R 2 , so the multiplication 
of the former produces (Tqf. 

If we put q = Sq + Vq, 

we shall have Kq = Sq Vq, 

and qKq = (Sq)' + (TVq)' t 

for (Vqf = -(TVqY, Art. 20. 



It is almost self-evident that, since the change of order of 
multiplication of two vectors produces no other change than that 
of the sign of the vector part of the product (22), 



q and r occurring in a changed order. 

The following is a demonstration, 

Let q=Tq( cos + a sin 0), 

r = Tr( cos < + ft sin <), 
a and ft being unit vectors ; then 

qr = TqTr (cos cos <f> a sin 6 cos < - (3 cos 6 sin < 

+ aft sin sin <), 

KrKq = TqTr (- cos < - (3 sin <) (- cos - a sin 0) 

= TqTr (cos 6 cos < + a sin 6 cos < + ft cos sin <f> 

+ fta sin sin <). 

Now observing that /3a has the same scalar part with aft, but 
the vector part with a contrary sign, we see that the two ex- 



5'4 QUATERNIONS. [CHAP. III. 

pressions for qr and for KrKq likewise have the same scalar 
part, but that their vector parts have contrary signs. 

Hence K (qr) = KrKq. 

(See Tait, 79 et sq.) 

31. We propose, in this Article, to give and interpret one or 
two formulae, relating to three or more vectors, which are indis- 
pensable to our progress, reserving to a separate Chapter the 
demonstration and application of other formulae, the value of 
which the reader can hardly as yet be expected to understand. 

1. To express S . a/3y geometrically. 

First suppose a, ft, y to be unit vectors OA, OB, 00. 

Let AOB-Q, and the angle which 00 makes with the plane 
A OB = (ft ; then since 

aft = - cos 6 + e sin 6 (Art. 21), 
where e is perpendicular to the plane A OB, 

S . afty = S(- cos + e sin 6) y 

= /Scy sin 6. 
Now Sey = cos . angle between e and y 

= sin . angle between plane AOB 

and 00 
= sin <f> 

.'. S. afty= sin <f) sin 0. 0< 

Next if a, ft, y are not units, but have re- 
spectively the lengths Ta, Tft, Ty, or a, b, c; 
we shall have 

S . afty abc sin sin $. 

But db sin is the area of the parallelogram of which the 
adjacent sides are a, b ', and c sin < is the perpendicular from C on 
the plane of the parallelogram ; 

. '. S . afty = db sin 6 . c sin < 

= volume of parallelepiped of which three con- 
terminous edges are OA, OB, 00. 




ART. 31.] VECTOR MULTIPLICATION AND DIVISION. 55 

2. From the nature of the case, no change of order amongst 

the vectors a, /?, y can make any change in the value (apart from 

the sign) of the scalar of the product of the three vectors ; for it 

will in every case produce the volume of the same parallelepiped. 

.-. S.aj3y = S .yap = S.ay/3, &c. 



COR. 1. The volume of the triangular pyramid, of which OA, 
OB, OC are conterminous edges is -^S . a/2y. 

COR. 2. If a, /?, y are in the same plane, </> = ; 
.'. S.ay = 0. 

Conversely, if S . a/?y = 0, none of the vectors a, /?, y being 
themselves 0, we must have either or <f> = ; hence in either 
case the three vectors are co-planar. 

3. Since Fa/3 = y' (21. 3), a vector perpendicular to the plane 
OAB (fig. of formula 2) ; F/?y = a', a vector perpendicular to 
the plane OBC ; and since y, a! are both perpendicular to 0, 
the line along which is the vector (3 ; OB is perpendicular to the 
plane which passes through y', a', and therefore (21. 3) is in the 
direction of Vy'a ; hence 

V ( Vafi Vfty) = Vy'a - mfi, 

or the vector of the product of two resultant vectors, one of the 
constituents of each of which is the same vectoi 1 , is a multiple of 
that vector. 

4. If OA = a, 0.8 = (3, OJ) = S, OE=e; and if the planes 
OAB, ODE intersect in OP; it follows, as in (3), that, Vafi and 
FSe being both perpendicular to OP, 

V(VapVSe) is along OP and is therefore =nOP. 

5. Connection between the representation of the position of a 
point by a vector and its representation by Cartesian co-ordinates. 

If a?, y, z be the perpendicular distances of a point P in space 
from the planes of yz, zx, xy respectively (fig. of Art. 16); *, j, k 



56 QUATERNIONS. [CHAP. III. 

unit vectors in the directions of x, y, z ; then xi is the vector of 
which the line is x (Art. 3) ; consequently OM along Ox, MN 
parallel to Oy and NP parallel to Os, being x, y, z as co-ordinates, 
they are xi, yj, zk as vectors. 

Now vector OP = OM+ MN+ NP, 

and is therefore p = xi + yj + zk. 

The same method of representation is evidently applicable 
when the planes of reference are not mutually at right angles. 
If x, y, z be the co-ordinates of P referred to oblique co-ordinates; 
a, /?, y unit vectors parallel respectively to x, y, z ; then 

vector OP = xa + y{3 + zy. 

COR. When x, y, z are at right angles to one another, 

p = xi + yj + zk 
gives Sip = -x, Sjp = -y, Skp = -z; 

.-. (Sip)' + (Sjp)> + (Sk P )* = x' + y* + z* 
= OP*. 

Ex. To find the volume of the pyramid of which the vertex is 
a given point and the base the triangle formed by joining three 
given points in the rectangular co-ordinate axes. 

Let A, B, C be the three given points ; 



x, y, z the co-ordinates of the given point P, 
then vector OA = ai, OB = bj, OC-ckj 

and OP = xi + 



= -{xi + yj+(z-c)k}. 



ART. 31.] VECTOR MULTIPLICATION AND DIVISION. 57 

Now the volume of the pyramid PASO is 

~S(PA.PB.PC} (31. 2. Cor. 1) 
= - ^ S . { (x - a) i + yj + zk} {xi + (y - b) j + zk} [xi + yk + (z - c) k}. 

Multiplying out and observing that only terms which involve 
all of the three vectors i, j, k produce a scalar in the product, 
we get 

(+ or -) Yol. = - ^ {(x - a) (bz + cy be) - cxy bxz} 



x y z n \ 
-+^ + 1). 
a o c J 



1 fx y z 

^A>. i i ^ i 

"6 

The sign of the result will of course depend on the position 
P. 



ADDITIONAL EXAMPLES TO CHAP. III. 

1. If in the figure of Euclid i. 47 DF, GH, KE be joined, 
the sum of the squares of the joining lines is three times the sum 
of the squares of the sides of the triangle. 

The same is true whatever be the angle A. 

2. Prove that 

AD* (Art. 7, Ex. 4) = 2 (AB* + AC*)- BO 3 . 

3. If P, Q, R, S be points in the sides AB, BC, 02), DA of 

a rectangle, such that PQ US, prove that 

AH* + OS' = AQ* + OP 2 . 

4. The sum of the squares of the three sides of a triangle is 
equal to three times the sum of the squares of the lines drawn 
from the angles to the mean point of the triangle. 



58 QUATERNIONS. [CHAP. III. 

5. In any quadrilateral, the product of the two diagonals and 
the cosine of their contained angle is equal to the sum or difference 
of the two corresponding products for the pairs of opposite sides. 

6. If a, b, c be three conterminous edges of a rectangular 
parallelepiped ; prove that four times the square of the area of 
the triangle which joins their extremities is 



7. If two pairs of opposite edges of a tetrahedron be respect- 
ively at right angles, the third pair will be also at right angles. 

8. Given that each edge of a tetrahedron is equal to the edge 
opposite to it. Prove that the lines which join the points of 
bisection of opposite edges are at right angles to those edges. 

9. If from the vertex of a tetrahedron OABG the straight 
line OD be drawn to the base making equal angles with the 
faces OAB, OAC, OBC ; prove that the triangles OAB, OAC, OBG 
are to one another as the triangles DAB, DAG, DBG. 



CHAPTER IV. 

THE STRAIGHT LINE AND PLANE. 

32. EQUATIONS of a straight line. 

1. Let ft be a vector (unit or otherwise) parallel to or along 
the straight line: a the vector to a given 

D A P 

point A in the line, p that to any point what- 
ever P in the line, starting from the same 
origin ; then AP is a vector parallel to /3 

= x{3, say, 

and OP = OA + AP 

gives p = a + x(3(I) 

as the equation of the line. 

2. Another form in which the equation of a straight line 
may be expressed is this : let A a, OS = (3 be the vectors to 
two given points in the line ; then 



Of course the ft of No. 2 is not that of No. 1. The first form 
of the equation supposes the direction of the line and the position 
of one point in it to be given, the second form supposes two points 
in it to be given. 

3. A third form may be exhibited in which the perpendicular 
on the line from the origin is given. 



60 QUATERNIONS. [CHAP. IV. 

Let OD perpendicular to AP = 8; then 



because OD is perpendicular to AP (22. 7) ; 

Le.S&p = C (3), 
where C is a constant. 

(Note. In addition to this we must have the equation of the 
plane of the paper, in which p is tacitly supposed to lie. This 
may be written as Sep = 0.) 

33. Equation of a plane. 

Let P be any point in the plane, OD perpendicular to the 
plane ; and let 

OD = S, OP = p- } 

then p - 8 = DP, 

which is in a direction perpendicular to OD ; 



or 



COR. 1. If SBp = C be the equation of a plane, 8 is a vector 
in the direction perpendicular to the plane. 

Con. 2. If the plane pass through 0, p can have the value zero, 
. '. SBp = is the equation. 

COR. 3. Since a vector can be drawn in the plane through Z>, 
parallel to any given vector in or parallel to the plane ; if ft be 
any vector in or parallel to the plane, SS/3 = 0. 

34. We proceed to exhibit certain modifications of the 
equations of a straight line and plane, and one or two results 
immediately deducible from the forms of those equations. 

1. To find the equation of a straight line which is perpen- 
dicular to each of two given straight lines. 

Let /?, y be vectors from a given point A in the required line, 
and parallel respectively to the given lines. 



ART. 34.] THE STRAIGHT LINE AND PLANE. 61 

If OA = a as before, then since (22. 8) F/3y LS a vector along 
the line whose equation is required ; we have 

p a = x T /7 /3y, 
or p = a + x F/3y, 
as the equation of the line. 

2. To find the length of the perpendicular from the origin on 
a given line. 

Equation (1) of Art. 32 is 

p - a. + x(3. 

If now p = OD = 8 ; 

we get S&* = SSa, 

or -OD* = SSa 



US being the unit vector perpendicular to the line. 
COR. The same result is true of a plane. 

3. To find the length of the perpendicular from a given point 
on a given plane. 

Let Sap = C be the equation of the plane, y the vector to the 
given point. 

Then if the vector perpendicular be xa (33. Cor. 1), 

p = y 4- xa 

gives Say + xa* = (7, 

and the vector perpendicular is 

xa = + a" 1 (C - Say) ; 

the square of which with a sign is the square of the perpendi- 
cular. 

4. To find the length of the common perpendicular to each 
of two given straight lines. 



62 QUATERNIONS. [CHAP. IV. 

Let (3, /3 l be unit vectors along the lines ; a, a x vectors to 
given points in the lines ; 

p = a + x(3, 

Pi = !+, A , 

the vectors to the extremities of the common perpendicular 8. 

Then since 8 is perpendicular to both lines, it is perpendicular 
to the plane which passes through two straight lines drawn pa- 
rallel to them through a given point ; 



But 8 = p - pj = a + x(3 aj - 

hence S . B/3/3, = S . (a - a,) 

i. e. S (y 7(3(3, . ft (3,) = S . (a - a,) 



because 

. 

whence 8 = /V is known. 



5. To find the equation of a plane which passes through three 
given points. 

Let a, ft, y be the vectors of the points. 

Then p a, a (3, (3 y are in the same plane. 

.-. (Art. 31. 2. Cor. 2) S. 0>-a)(a-/8)(-y) = 0, 
or Sp(Va(3+V(3y+Vya)-S.a(3y = 

is the equation required. 

COR. Fa/3 + V(3y + Vya is a vector in the direction perpen- 
dicular to the plane; therefore (No. 3) the perpendicular vector 
from the origin 

= S.a(3y.(Va(3+ V(3y + Fya)' 1 . 

6. To find the equation of a plane which shall pass through 
a given point and be parallel to each of two given straight lines. 



ART. 34.] THE STRAIGHT LINE AND PLANE. 63 

Let y be the vector to the given point, p = a + xft, p = a l + a; 1 /8 1 
the lines ; then if lines be drawn in the required plane parallel to 
each of the given straight lines these lines as vectors will be 
ft, ft 1 : also p y is a vector line in the plane ; 

.'. S.ftft 1 (p-y) = Q (31. 2. Cor. 2), 
which is the equation required. 

7. To find the equation of a plane which shall pass through 
two given points and be perpendicular to a given plane. 

Let a, ft be the vectors to the given points, SSp C the equa- 
tion of the plane ; then the three lines p a, a ft, 8 are vectors 
in the plane ; 



or .pa 

8. To find the condition that four points shall be in t/te same 
plane. 

1. Let OA, OS, 00, OD or a, ft, y, 8 be the vectors to the 
four points ; then 8 a, 8 ft, 8 y are vectors in the same plane ; 

.-. S . (8 - a) (8 - ft) (8 - y) = (31. 2. Cor. 2), 
or S.ofty + S.a$y + S.aftS = S.afty (1). 

2. Another form of the condition is to be obtained by as- 
suming that 

dS + cy + bft + aa = (2), 

and substituting in equation (1) the value of 8 deduced from 
this equation. The result is 

a o c - - 



= Q (3). 

Equation (1), or the concurrence of equations (2) and (3) is the 
condition necessary and sufficient for coplanarity. 

9. To find the line of intersection of two planes through the 
origin. 



64 QUATERNIONS. [CHAP. IV. 

Let Sap = 0, Sftp = be the planes. 

Since every line in the one plane is perpendicular to a ; and 
every line in the other perpendicular to ft; the line required is 
perpendicular to both a and ft, and is therefore parallel to Fa/?, 
or p = xVaft is the equation. 

10. The equation of the plane which passes through and 
the line of intersection of the planes /Sap = a, Sftp = b is 

Sp(aft-ba) = 0. 

For 1 it is a plane through ; 2 if p be such that Sap = a, 
then must Sftp = b. 

11. To find the equation of the line of intersection of the two 
planes. 

Let p = ma + nft + xVaft 

be the equation required. 

Then Sap = ma? + nSaft = a, 

since Vaft is perpendicular to a, and similarly 



aft*-bSaft bSaft-aft* 
~ a*ft*-(Saft)* ~ (Vaft)* 

aSaft-ba 2 aSaB-ba* 



(Saft)*-a 2 ft 2 ~ (Vaft) 1 
35. We offer a few simple examples. 



Ex. 1. To find the locus of the middle points of all straight 
lines which are terminated by two given straight lines. 

Let AP, BQ be the two given straight 
lines, unit vectors parallel to which are ft, y; 
AB the line which is perpendicular to both 
AP, BQ. 

Let be the middle point of AB; vector 
A = a ; R the middle point of any line PQ, 
rector OR = p ', then 




ART. So.] THE STRAIGHT LINE AND PLANE. 65 



But 



hence, since Sa/3 = 0, Say = 0, 

Sap = is the equation required ; and the locus is a plane passing 
through (33. Cor. 2), and perpendicular to OA (33. Cor. 1). 

Note that, if /? || y, we have simply 
2p = x'(3; 

and, as there is now but one scalar indeterminate, the locus is a 
straight line instead of a plane. 

Ex. 2. Planes cut off, from, the three rectangular co-ordinate 
axes, pyramids of equal volume, to find the locus of the feet of per- 
pendiculars on them from the origin. 

Here the axes are given, so that i,j, k are known unit vectors. 

Let ai, bj, ck be the portions cut off from the axes by a plane, 
the perpendicular on which from the origin is p. 

Then p ai is perpendicular to p ; 



or p = 
Similarly, p 2 

p 2 = cSkp. 

Hen ce p 6 = abc Sip Sjp,Skp 

= CSipSjpSkp, 
since abc is by the problem constant. 

If x, y, z be the co-ordinates of p this equation gives at once 



as the equation required. 
T. Q. 



66 QUATERNIONS. [CEAP. IV. 

Ex. 3. To find the locus of the middle points of straight lines 
terminated by two given straight lines and all parallel to a given 
plane. 

Retaining the figure and notation of Ex. 1, let 8 be the vector 
perpendicular to the given plane : we have 



Now SBQP = (33. Cor. 3); 



2SoS S/3S 

and 2p = xp + r-^ y + x ~~ y 

oyo oyo 



where a ~ , 6 = ^;- are constants; (oyd for instance is the 
oyo oyo 

negative of the cosine of the angle between one of the given lines 
and the perpendicular to the given plane). 

Now (B + by is a known vector lying between /3 and y ; call it 
e, and 2p = ay + xe is the equation required; which is that of a 
straight line, not generally passing through (32. 1). 

Ex. 4. OA, OB are two fixed lines, which are cut by lint's 
AS, A'B' so tJutt the area AOJB is constant/ and also the product 
OA, OA' constant. It is required to find the locus of the intersec- 
tions of AS, A'B'. 

Let the unit vectors along OA, OB be a, ft respectively. 
OA = ma, OA' m'a, 



then the conditions of the problem are 
mn = m'n' = C, 
mm' = a. 



ART. 35.] THE STRAIGHT LINE AND PLANE. 67 

Now if A, A'B' intersect in P, and OP = p, we have 
P =OA + AP 

= ma + x (nj3 ma), 
p = OA' + A'P 

m'a + x' (rift m'a) ; 

or p = ma + xl B 

/G \ 

p = m'a + x' ( -. B m'a } ; 
\m r J 

x x' 

and = > . 

m m 

AM 

Hence x = 



m + m 



m + 
a 



and p = 5 (aa + CS), 

m* + a^ 

and the locus required is a straight line, the diagonal of the 
parallelogram whose sides are aa, Cf$. 

Ex. 5. To find the locus of a point such that tlie ratio of its 
distances from a given point and a given straight line is constant 
all in one plane. 

Let S be the given point, DQ the given 
straight line, SP = ePQ the given relation. 

Let vector SD = a,SP = p, DQ = yy, 
y being the unit vector along DQ, 

PQ = xa; 
then T P = eT(PQ), 

52 



68 QUATERNIONS. [CHAP. IV. 

gives p 2 = e 2 PQ 2 , where PQ is a vector, 

= e* (xa)" 



But 



. \ Sap + xa a = a 2 , for Say = ; 
and a?of=(cL*-Sap) a ; 

hence a'p' = e* (a 2 - Sap) 2 , 

a surface of the second order, whose intersection with the plane 
S . ayp = is the required locus. 

Ex. 6. TJie same problem when the points and line are not in 
the same plane. 

Retaining the same figure and notation, we see that PQ is no 
longer a multiple of a ; but 



and because PQ is perpendicular to DQ 



and p 9 = e 2 (a ySyp p) 2 , 

a surface of the second order. 

COR. If e = 1, and the surface be cut by a plane perpendicular 
to DQ whose equation is Syp = c, the equation of the section is' 



another plane, so that the section is a straight line. 

Ex. 7. To find the locus oftJie middle points of lines of given 
length terminated by each of two given straight lines. 



ART. 35.] THE STRAIGHT LINE AND PLANE. 



69' 



Retaining the figure and notation of Ex. 1, and calling RP c, 
we have 

2p = xp + yy (1), 

and 2RP = RP-JtQ=2a + xj3-yy (2). 

From equation (1) we have 

Sap = (22. 7), 



because (3 is a unit vector, 

2Syp = xS(3y y. 

The first of these three equations shews that p lies in a plane 
through perpendicular to AB (33. Cor. 2). 
The second and third equations give 

2(Sf3p+S{lySyp) 



Now (2) gives, by squaring, 
- 4c 2 = 4a 2 + x*( 

in which, if the values of x and y just obtained be substituted, 
there results an equation of the second order in p. 

Hence the locus required is a plane curve of the second order, 
or a conic section, which by the very nature of the problem must 
be finite in extent and therefore an ellipse. 

Ex. 8. If a plane be drawn through the points of bisection of 
two opposite edges of a tetrahedron it will bisect the tetrahedron. 

Let D, E be the middle points of OB, 
AC: DFEG the cutting plane: OA, OB, 
OC = a, ft, y respectively. 

OG = my, AF=n((3-a}. 

The portion ODGEA consists of three 
tetrahedra whose common vertex is 0, and 
bases the triangles AEF, EFG, FGD. 

Kow OE= l - + a 




70 QUATERNIONS. [CHAP. IV 

00 -I* 

OG=my, 

OF=a + n(p-a); 

and 6 times the volume cut off 



+ S. TJ (a + y) my {a + n ((3 - a)} 



P (31.2 Cor. 



--{n + nm + (1 - n) m} S . ay ft 



. ay/3. 

But since E, G, D, F are in one plane, and 

2m (1 - ) OE - (1 - n) OG + 2mnOD - mOF= 0, 
we must have (34. 8) 

2m (1 - n) - (1 - n) + 2mn - m = ; 

.'. m + n = 1 ; 
and 6 times the whole volume cut off 



= jr of 6 times the whole volume, 
t 

hence the plane bisects the tetrahedron. 

COR. The plane cuts other two edges at F and G, so that 

AF_ OG_ 
AE + OC 



EX. 1.] THE STRAIGHT LINE AND PLANE. 71 

ADDITIONAL EXAMPLES TO CHAP. IV. 

1. Straight lines are drawn terminated by two given straight 
lines, to find the locus of a point in them whose distances from 
the extremities have a given ratio. 

2. Two lines and a point S are given, not in one plane ; find 
the locus of a point P such that a perpendicular from it on one 
of the given lines intersects the other, and the portion of the 
perpendicular between the point of section and P bears to SP 
a constant ratio. Prove that the locus of P is a surface of the 
second order. 

3. Prove that the section of this surface by a plane perpen- 
dicular to the line to which the generating lines are drawn pei'pen- 
dicular is a circle. 

4. Prove that the locus of a point whose distances from two 
given straight lines have a constant ratio is a surface of the second 
order. 

5. A straight line moves parallel to a fixed plane and is ter- 
minated by two given straight lines not in one plane ; find the 
locus of the point which divides the line into parts which have 
a constant ratio. 

6. Required the locus of a point P such that the sum of the 
projections of OP on OA and OB is constant. 

7. If the sum of the perpendiculars on two given planes from 
the point A is the same as the sum of the perpendiculars from B, 
this sum is the same for every point in the line AB. 

8. If the sum of the perpendiculars on two given planes from 
each of three points A, B, C (not in the same straight line) be the 
same, this sum will remain the same for every point in the plane 
ABC. 

9. A solid angle is contained by four plane angles. Through 
a given point in one of the edges to draw a plane so that the sec- 
tion shall be a parallelogram. 



72 QUATERNIONS. [CHAP. IV. 

10. Through each of the edges of a tetrahedron a plane is 
drawn perpendicular to the opposite face. Prove that these planes 
pass through the same straight line. 

11. ABC is a triangle formed by joining points in the rect- 
angular co-ordinates OA, OB, OC ; OD is perpendicular to ABC. 
Prove that the triangle AOB is a mean proportional between the 
triangles ABC, ABD. 

12. VapVfip + ( Fa/3) 2 = is the equation of a hyperbola in p, 
the asymptotes being parallel to a, (3. 



CHAPTER V. 

THE CIRCLE AND SPHERE. 

36. Equations of the circle. 
Let AD be the diameter of the circle, 
centre (7, radius = a, P any point. 

If vector CD = a, CP = p, 

we have p 2 = a 2 (1). A 

If however AP = p, 



we have (p-a) 2 = -a 2 
If be any point, . 



(2). 




we have (p~y) 2 ? (3)- 

These are the three forms of the vector equation. 
Form (2) may be written 



If OC = c, form (3) may be written 



-a. 

EXAMPLES. 

37. Ex. 1. Tlve angle in a semicircle is a right angle. 
Taking the second form 

p 3 - 2Sap = 0, 
we may again write it 



74- QUATERNIONS. [CHAP. V. 

therefore p, p 2a are vectors at right angles to one another. 
But p - 2a is DP ; 

.. DP A is a right angle. 

Ex. 2. If through any point within or ivithout a circle, a 
straight line be drawn cutting the circle in the points P, Q, the pro- 
duct OP . OQ is always the same for that point. 

The third form of the equation may be written, 

(TpY + 2TpS 7 Up + c* - a? = 0, 

which shews that Tp has two values corresponding to each value 
of Up, the product of which is c 2 a 2 . Therefore, &c. 

Ex. 3. If two circles cut one another, the straight line which 
joins the points of section is perpendicular to tJte straight line which 
joins the centres. 

Let 0, C be the centres, P, Q the points of section ; 

vector OC = a.' } a, b the radii; 
then (as vectors) 



.: iSa.OP = C, a constant. 

Similarly, SaOQ = C, the same constant ; 

.-. Sa(OQ-OP) = Q > 

or SaPQ = Q, 
i.e. PQ is at right angles to 00. 

Ex. 4. in a fixed point, A B a given straight line. A point Q 
is taken in the line OP drawn to a point P in AB, such that 

OP.OQ = k*-, 
to find the locus of Q. 

Let OA perpendicular to AB be a, vector a ; 

OQ = P , OP = xp; 

then T(OP.OQ) = k 2 , 

or xp 2 = -tf. 



ART. 37.] THE CIRCLE AND SPHERE. 75 

But So. (xp-a) = Q; 

.. xSap = a*; 

k* 

hence p 2 = -~ Sap 

a 

is the equation of the locus of Q, which is therefore a circle, 
passing through 0. 

Ex. 5. Straight lines are drawn through a fixed point, to find 
the locus of the feet of perpendiculars on them from another fixed 
point. 

Let 0, A be the points, the lines being drawn through A. 
Let OA a, and let p = a + x(3 be the equation of one of the lines 
through A, 8 the perpendicular on it from 0. 

Then 8 = a + xfi, 

and S8 3 = SaS, 

because 8 is perpendicular to ft ; 

i.e. o*-SaS = 0, 
the equation of a circle whose diameter is OA. 

Ex. 6. A chord QR is drawn parallel to the diameter AB of 
a circle : P is any point in AB ; to prove that 

PQ* + PR* = PA* + PB*. 

Let CQ = P , CR = p, PC = a; 

then PQ* = - (vector PQ)' 

= -(a. + p}* = - (a 3 + 2Sap + p 3 ), 
PR* = - (a + p') 3 = - (a 2 + 2Sap' + p' 2 ) ; 
.-. PQ* + PR 2 = 2PC 2 + 2AC'-2 (Sap + Sap'). 
But S(p + p') (p - p) = and p p' = xa, 

because QR is parallel to AB \ 

. '. Sap + Sap' = 0, 
and PQ 2 + PR 2 = 2PC 2 + 2 AC 3 



7G QUATERNIONS. [CHAP. V. 

Ex. 7. If three given circles be cut by any other circle, the 
chords of section will form a triangle, the loci of the angular points 
of which are three straight lines respectively perpendicular to the 
lines which join the centres of the given circles ; and these three 
lines meet in a point. 

Let A, B, C be the centres of the three given circles ; a, b, c 
their radii ; a, /?, y the vectors to A, B, C from the origin ', 
OA, OB, 00 respectively p, q, r ; D the centre of the cutting 
circle whose radius is R, OD = s, vector OD = 8, p the vector to 
a point of section of circle D with circle A ; then we shall have 



and .-. 

Now this is satisfied by the values of p to both points of sec- 
tion ; and being the equation of a straight line (32. 3) is the 
equation of the line joining the points of section of circle D with 
circle A call it line 1, and so of the others; then 

line 1 is 2S (8 - a) p = R 3 - a 2 - s* +p 2 , 
line 2 is 2S(& -p) P ' = R 2 -b*-s* + q 2 , 
line 3 is 2S(S - y) p" = H 2 -c 2 -s 2 + r 3 . 

If 1 and 2 intersect in P whose vector is p lt 1 and 3 in Q (p 2 ); 
2 and 3 in R (p a ), we shall have by subtraction 

atP, 



therefore (32. 3) the lod of P, Q, JR are straight lines, perpen- 
dicular respectively to AH, AC, BC. 

Also at the point of intersection of the first and third of these 
lines, we have, by addition, 



which is satisfied by the second : hence the three loci meet in a 
point. 



ART. 37.] THE CIRCLE AND SPHERE. 77 

Ex. 8. To find the equation of the cissoid. 

AQ is a chord in a circle whose diameter is AB, QN perpen- 
dicular to AB. 

AM is taken equal to BN, and MP is drawn perpendicular 
to A B to meet AQ in P ; the locus of P is the cissoid. 

Let vector AP = TT, AC = a, AM=ya, AQ = XTT; 
then y : 1 :: 2-y : x, by the construction ; 



Now 

is the equation of the circle ; 

2SW 

M _ 

7T* ' 

Also Tr 



a7r O.TT 

hence I 1 H -- 5- ) 5- = *i 

V TT / a 

and (7T 2 + 2^a7r) Sair = 2aV, 

is the equation required. 

Ex. 9. If ABCD is a parallelogram, and if a circle, be de- 
scribed passing through the point A, and cutting the sides AS, AC 
and the diagonal AD in the points F, G, H respectively ; then the 
rectangle AD . AH is equal to the sum of the rectangles AS . AF, 
andAG.AG. 



Let 



AF=xa, 



78 QUATERNIONS. [CHAP. V. 

6 the vector diameter of the circle ; then 



whence, since y = a + {3, 

zy* = xa* + yft 3 ; 
i.e. AD. AH = AB.AF+AC.AG. 



Ex. 10. What is represented by the equation 



If a, ft be not at right angles to one another, we can put 
a l + eft for a, and so choose e that Sa^ft = 0. 

We shall therefore consider a, ft as vectors at right angles 
to each other, and we may, on account of x, assume their tensors 
equal, and each a unit. 

a+xft a+xft 



Hence p = 

or, if I 



(a + xft)' ' 1+a?' 



p = - sin (a sin + ft cos 6), 
whence Tp (= r) = sin 0, 

a circle of which the diameter is a unit parallel to a and the 
origin a point in the circumference; and ft a tangent vector at 



the origin. 



Otherwise, Sap = 



or p 8 = Sap. 



AET. 38.] THE CIRCLE AND SPHERE. 79 

Or, again, p" 1 = a + x{3 ; 

whence Sap' 1 = 1 , 

or VP (p~ l - a) = 0, 



where U stands for the versor of the quaternion ; 
all of these being, with the obvious condition S . a ftp = 0, varieties 
of the form of the equation of a circle, referred to a point in the 
circumference, the diameter through which is parallel to a. 
Draw any two radii p and p t , then we have 

S. U ' 






PiP 

Pl ' P 7 2 P wiu be rendered a unit if we take a unit 

PiP 
vector along each of the three vectors p 1? (p - p ; ), and p ; 

.-. s. U 



But 



and S. Up~ l U (p~ l - p- 1 ) 

Hence S. Up,U(p~ Pl } =-S/3Up. 

If p be constant whilst p l varies, the right-hand side of this 
equation is constant, and the equation shews that the angles in 
the same segment of a circle are equal to one another. 

Further, the form of the right-hand side of the equation, viz. 
SfiUp, shews that the angle in the segment is equal to the sup- 
plement of the angle between the chord (p) and the tangent (/?). 

38. To draw a, tangent to a circle. 

1. If we assume the first form of the equation, the centre 
being the origin, and assume also that the tangent is at right 



80 QUATERNIONS. [CHAP. V. 

angles to the radius drawn to the point of contact ; we shall have, 
denoting by TT a vector to a point in the tangent, 

Sp (?r - p) = 0, 
for TT p is along the tangent ; 

. \ Sirp a 2 
is the equation required. 

2. Without assuming the property of the tangent, we may 
obtain it as follows. 

Let p be a point in the circle near to P ; then 

from the equation ; 

But p' 4- p is the vector which bisects the angle between the 
vectors to the points of section, and p p is a vector along the 
secant. 

Now the equation shews (22. 7) that the former of these lines 
is perpendicular to the latter. 

As the points of section approach one another, the tangent 
approaches the secant, and the bisecting line approaches the radius 
to the point of contact : therefore the radius to the point of 
contact is perpendicular to the tangent. 

39. From a point without a circle two tangents are drawn 
to the circle, to find the equation of the chord of contact. 

Let /3 be the vector to the given 
point, 

<? ~a / / \\ 

I ~ ^ / I \ n 

the equation of a tangent; then since 
it passes through the given point 




Now this equation is satisfied for both points of contact, and 
since it is the equation of a straight line (32. 3) it must be satis- 
fied for every point in the straight line which passes through those 
points : it is therefore the equation of the chord of contact. To 



ART. 40.] THE CIRCLE AND SPHERE. 81 

avoid the appearance of limiting p to a point in the circle, we may 
write a- -in place of p ; and the equation of the chord of contact 

becomes 

Spa- = - a 3 . 



EXAMPLES. 

40. Ex. 1. If chords be drawn through a given point, and 
tangents be drawn at the points of section, the corresponding pairs 
of tangents will intersect in a straight line. 

Let y be the vector to the given point G, the centre C being 
the origin ; ft the vector to 0, the point of intersection of two 
tangents at the extremities of a chord through G ; then the equa- 
tion of the chord of contact is (39) 

S/3<r=-a s , 
and as the chord passes through G we have 



which, since y is a constant vector, is the equation of a straight 
line, the locus of ft. 

COR. 1. The straight line is at right angles to CG (32. 3). 

COR. 2. The converse is obviously true, that if through points 
in a straight line pairs of tangents be drawn to a circle, the chords 
of contact all pass through the same point. 

Ex. 2. Any chord drawn from the point of intersection of 
two tangents, is cut harmonically by the circle and the chord of 
contact. 

Let radius = a, 0(7 -c, OR=p, OS=q, vector OC=a, unit 
vector OR = p ; then 



is the equation of the circle ; 

i.e. p 1 + 2pSap + c" - a 3 0, 
T. Q. 



QUATERNIONS. 



[CHAP. v. 



a quadratic equation which gives 
the two values of p, viz. OR and 
OT; 

JL _L. 2Sa P 

' 777? O7 r ~ /* // 2 " 

L/.Zli \s J- C tw 



Saqp = SaON; 




hence 



2__2 
0^~g 



2Sap 



_L ! 

~0/i + OT' 

Ex. 3. 7/" tangents be drawn at the angular points of a triangle 
inscribed in a circle, the intersections of these tangents with the 
opposite sides of the triangle lie in a straight line. 




Let radius = a, OA = a, OE = fi, 00 = 7, then 



ART. 40.] THE CIRCLE AND SPHERE. 

But a is perpendicular to AP ; 



83 



a' + Sap 
Say -Sap' 



and 



Say - Sap 



oap opy 



Sim ii arly , 



^ 
Say 

Hence (Say - Saft) OP + (Sap - Spy) OQ 

+ (Spy -Say) OB = 0, 
whilst (Say - Sap) + (Sap - Spy) + (Spy - Say) = 0. 

Consequently (Art. 13) P, Q, R are in the same straight line. 

COR. PQ : PR :: Spy-Say :: Spy-S a p 

:: cos 2.C cos 2 J. : cos 2(7 cos 2A 
:: sin C sin (B - A) : 



Ex. 4. A faced circle is cut by a number of circles, all of which 
pass through two given points ; to prove that the lines of section of 
the fixed circle with each circle of the series all pass through a point 
whose distances from the two given points are proportional to the 
squares of the tangents drawn from those points to the fixed circle. 

Let be the centre of the 
fixed circle whose radius is a, 
A, B the given points, vectors 
a, p, the origin being ; OA = b, 
OB c;C the centre of a circle 
which passes through A and B, 
radius r 00 = p, TT the vector to 
any point in the circumference of 
this circle; then the equation of 
the circle is (TT p) 2 = r 2 ; 

62 




84 QUATERNIONS. [CHAP. V. 

hence for the four points A, JB, P, Q, we have 
a 2 - 2Sap + p 2 = - r\ 



From which it follows that 

S(OP-OQ) P = ....................... (1), 

-b* + c 2 = a:-p z = 2S(a-p)p ................ (2), 

2S(OP-a)p=OP 2 -a 2 = -a i +b 2 ............... (3). 

Let QP, AB intersect in R, OR = a- ; then 



= S.OP P \y(l), 

and Sa-p = S {a + y (a - /?)} p 



= 2S(OP-a)p 
=-sf + V\ty(S), 

i.e. y is independent of p and r ; or R is the same point for 
every circle : 

(c'-tt')a-(6'-a')0 

also OR = - - 1-3 ^ - H- , 

c o 

and RA : RB :: a- OR : ft- OR :: V-a* : c*-a* 

:: AT 2 : BU*. 

41. The Sphere. 

1. It is clear that there is nothing in the demonstration of 
Art. 36 which limits the conclusions to one plane ; it follows that 
the equations there obtained are also equations of a sphere. 

2. Further if we assume that the tangent plane to a sphere 
is perpendicular to the radius to the point of contact, the con- 
clusion in Art. 38 is applicable also. 



ART. 42.] THE CIRCLE AND SPHERE. 85 

The equation of the tangent plane to the sphere is therefore 



3. Lastly, the results of Art. 39 are also applicable if we 
substitute any number of tangent planes passing through a given 
point for two tangent lines ; the equation of the plane which 
passes through the points of contact is therefore 

S/3<r=-a*. 

This plane is the polar plane to the point through which the 
tangent planes pass. 

COR. Since the polar plane is perpendicular to the line which 
joins the centre with the point through which the tangent planes 
pass, the perpendicular CD to it from the centre is along this 
line and has therefore the same unit vector with it. The equa- 
tion above gives in this case 



.-. CO. CD = a 2 (19). 

EXAMPLES. 

42. Ex- ! Every section of a sphere made by a plane is 
a circle. 

Let p 2 = a 8 be the equation of the sphere, a the vector per- 
pendicular from the centre on the cutting plane ; c the correspond- 
ing line. 

Let p = a + TT ; 

then the equation becomes 



But Sair = ; 

.-. 7r 2 = -(a 2 -c 2 ) 

is the equation of the section, which is therefore a circle, the square 
of whose radius is a 2 c 2 . 

Ex. 2. To find the curve of intersection of two spheres. 
Let the equations be 



p 2 -2Sa.' P =C'; 



86 QUATERNIONS. [CHAP. V. 

.'. 2S(a'-a)p=C-C', 

a plane perpendicular to the line of which the vector is a' a, 
i.e. to the line which joins the centres of the two spheres. 
Hence, by Ex. 1, the curve of intersection is a circle. 

Ex. 3. To find the locus of the feet of perpendiculars from the 
origin on planes which pass through a given point. 

Let a be the vector to the point, 8 perpendicular on a plane 
through it ; then 



is the equation of that plane ; therefore for the foot of the per- 
pendicular 

S(S 2 -aS)=0; 

or S 2 -SaS = 

is true for the foot of every perpendicular and is therefore the 
equation of the surface required. Hence it is a sphere whose 
diameter is the line joining the origin with the given point. 

Ex. 4. Perpendiculars are drawn from a point on the surface 
of a sphere to all tangent planes, to find the locus of their extremi- 
ties. 

Let a be the vector to the given point, 

Sirp = a" 
the equation of a tangent plane. 

Since the perpendicular is parallel to p, its vector is 
TT = a + xp ', 



because both p and a are vector radii. 
But Sirp = a* gives with xp = TT - a, 
STT (IT a) = a*x, 
(** - Sair)* = a*x s 

= a 2 x a*x' 
= -a a (7r-a) f . 



ART. 42.] THE CIRCLE AND SPHERE. 87 

Ex. 5. If the points from which tangent planes are drawn to 
a sphere lie always in a straigM line, prove that the planes of sec- 
tion all pass through a given point. 

Let CE be perpendicular to the line in which the point ft 
lies (41), see fig. of Art. 39, 

CE=c, vector CE=8; 
then SpS = -c> 

is the equation of the line. 

But Sp<r = -a* 

is the plane of contact, which is therefore satisfied by 



i. e. the planes all pass through a point G in CE, such that 

CG = - a CE, 
or CE.CG=a\ 

Ex. 6. If three spheres intersect one another, their three planes 
of intersection all pass through the same straight line. 

Let a, /?, y be the vectors to the centres of the three spheres, 
p'-2Sap=a, 



their three equations ; 
.-. 

2S (a y) p = c - a, 



are the equations of the three planes of intersection. 

Now the line of intersection of the first and second of these 
planes is obtained by taking p so as to satisfy both equations, 
and therefore their difference 



88 QUATERNIONS. [CHAP. V. 

which, being the third equation, proves that the same value of p 
satisfies it also. The three planes consequently all pass through 
the same straight line. 

Ex. 7. To find tlie locus of a point, the sum of the squares 
of whose distances from a number of given points has a given 
value. 

Let p denote the sought point ; a, /?,... the given ones ; then 



If there be n given points ; this is 



or 



A>_! a y = (^y_!(2.a 2 +(7). 

\ r n } \n J n^ 



This is the equation of a sphere, the vector to whose centre is 

-2 (a), 
n 

i. e. the centre of inertia of the n points taken as equal. 
Transpose the origin to this point, then (36) 



and /' = - {* (a f ) + 0}- 

Hence, that there may be a real locus, C must be positive 
and not less than the sum of the squares of the distances of the 
given system of points from their centre of inertia. If C have 
its least value, we have of course 

*> = 0, 
the sphere having shrunk to a point. 

ADDITIONAL EXAMPLES TO CHAP. V. 

1. If two circles cub one another, and from one of the points 
of section diameters be drawn to both circles, their other extre- 
mities and the other point of section will be in a straight line. 



EX. 2.] ADDITIONAL EXAMPLES. 89 

2. If a chord be drawn parallel to the diameter of a circle, 
the radii to the points where it meets the circle make equal angles 
with the diameter. 

3. The locus of a point from which two unequal circles sub- 
tend equal angles is a circle. 

4. A line moves so that the sum of the perpendiculars on it 
from two given points in its plane is constant. Shew that the 
locus of the middle point between the feet of the perpendiculars 
is a circle. 

5. If 0, 0' be the centres of two circles, the circumference 
of the latter of which passes through ; then the point of inter- 
section A of the circles being joined with 0' and produced to 
meet the circles in G, D, we shall have 



6. If two circles touch one another in 0, and two common 
chords be drawn through at right angles to one another, the 
sum of their squares is equal to the square of the sum of the 
diameters of the circles. 

7. A, , G are three points in the circumference of a circle; 
prove that if tangents at E and G meet in D, those at C and A 
in E, and those at A and B in F; then AD, BE, CF will meet 
in a point. 

8. If A, B, G are three points in the circumference of a 
circle, prove that V (AB . BC . CA) is a vector parallel to the tan- 
gent at A. 

9. A straight line is drawn from a given point to a point 
P on a given sphere : a point Q is taken in OP so that 

OP.OQ^k 3 . 
Prove that the locus of Q is a sphere. 

10. A point moves so that the ratio of its distances from two 
given points is constant. Prove that its locus is either a plane 
or a sphere. 



90 QUATERNIONS. [CHAP. V. 

11. A point moves so that the sum of the squares of its 
distances from a number of given points is constant. Prove that 
its locus is a sphere. 

12. A sphere touches each of two given straight lines which 
do not meet ; find the locus of its centre. 



CHAPTER VI. 

THE ELLIPSE. 

43. ! I* 1 "we define a conic section as "the locus of a point 
which moves so that its distance from a fixed point bears a con- 
stant ratio to its distance from a fixed straight line " (Todhunter, 
Art. 123), we shall find the equation to be (Ex. 5, Art. 35) 

ay = e 8 (a 2 -Sap) 2 (1), 

where SP = ePQ, vector SD = a, SP = p. 

When f is less than 1, the curve is the ellipse, a few of whose 
properties we are about to exhibit. 

2. SA, SA' are multiples of a : call one of them xa : then, 
by equation (1), putting xa for p, we get 




92 QUATERNIONS. [CHAP. VI. 



+e 



.-. AA'=^ 9 SD, 

J. ~~ 6 

the major axis of the ellipse, which we shall as usual abbreviate 
by 2a. 

If C be the centre of the ellipse 



J.-e 1e* 

= ae, 

and if vector CS be designated by a, CP by p, we have 

I a, and p' p + a' ; 



L 

whence, by substituting in (1), the equation assumes the form 

ay 2 +(&*>')' = - a 4 (1-e 2 ); 
which we may now write, CS being a and CP p, 

-a 4 (l-e 2 ) .................... (2). 



3. This equation might have been obtained at once by re- 
ferring the ellipse to the two foci, as Newton does in the Prin- 
cipia, Book i. Prop. 1 1 ; the definition then becomes 



or in vectors, if 



i. e. J- (p + a) i + J-(p- a)" = 2a ; 
hence, squaring, 

a J (p a) 2 = a 3 + Sap ; 



ART. 45.] THE ELLIPSE. 93 

If now we write <t>p for --- , where <4p is a vector 

a (1 - e) 

which coincides with p only in the cases in which either a coin- 
cides with p or when Sap = 0, i. e in the cases of the principal 
axes ; the equation of the ellipse becomes 



1 ............................... (3). 

The same equation is, of course, applicable to the hyperbola, 
e being greater than 1. 

44. The following properties of <f>p will be very frequently 
employed. The reader is requested to bear them constantly in 
mind. 

1. < (p + <r) = <f>p + (fur. 
= X(f>p. 

a 2 S<rp 4- SarrSap 







a* (1-0 



They need no other demonstration than what results from 
simple inspection of the value of <p 

a'p + aSap 

~ *(i-o ' 

45. To find the equation of the tangent to the ellipse. 

The tangent is defined to be the limit to which the secant 
approaches as the points of section approach each other. 

Let CP = p, CQ = p f , then 

vector PQ = CQ - GP = p - p = ft say ; 
j8 is therefore a vector along the secant. 

Now Sp'<t>p =S( P + P)<f>(p + P) 

4>P) (44.1) 



QUATERNIONS. [CHAP. VI. 

But Sp'fo' = 1 = Sp<f>p 5 



or (44. 3) 2Sp<f>p + SP<I>P = 0. 

Now P<f>p involves the first power of P whilst P^p involves 
the second, and the definition requires that the limit of the sum 
of the two as P gets smaller and smaller should be the first only, 
even if that should be zero : i. e. when P is along the tangent, we 
must have 

= 0. 



[We might also have written the equation in the form 

lff.0(*H-3tf),-* 

Thus, however small the tensor of P may be, 



is always perpendicular to /?. Whence, finally, 

/?&> = 0.] 
Let then T be any point in the tangent, vector CT = IT, then 

it = p + xp, 
and Sp<f>p = gives 



. '. Sir<f>p = Sp(j>p = 1 
is the equation of the tangent. 

COR. 1 <}>p is a vector along the perpendicular to the tangent 
(32. 3), that is, <f>p is a normal vector, or parallel to a normal 
vector at the point p. 

COE. 2. The equation of the tangent may also be written 
(44. 3) Spfir = 1. 



46. We may now exhibit the corresponding equations in 
terms of the Cartesian co-ordinates, as some of the results are 
best known in that form. 



ART. 46.] THE ELLIPSE. 95 

Let CM=x, MP y as usual; then, retaining the notation 
of Art. 31 with i, j as unit vectors parallel and perpendicular 
respectively to CA, 

vector CM- xi, HP = yj, OS - aei ; 
.-. p = xi + yj, 

a*p + aSap 
* p= "a*(l-O 

a* (1 e*) xi + cfyj 



, 30 
'tf+ 

where 6 2 -a s l-e s ; 

and 



. L 4.2-ssl 

' a 3 b'~ 
is the Cartesian interpretation of Sp<f>p = 1. 

Again, if x', y be the co-ordinates of T a point in the tangent, 



and S*<t> P = -S (x'i + y'j] + 



is the equation of the tangent. 



96 QUATERNIONS. [CHAP. VI. 

47. The values of p and <j>p exhibited in the last Article, 



viz. 



enable us to write 



a 
We shall have 



, t 



, . 
= 9<PP = 



If, further, we write \}/p for 



+ 
we shall have 



l p = (aiSip 4- bjSjp), &c. 

P = 'TVp 

(5). 



It is evident that the properties of <f>p (Art. 44) are possessed 
by all these functions. 



Now 

gives Sp\j/ (if/p) --1. 

But since Sptyo- = 

this becomes fyp^P = 

or Ttyp = 1 ; 



ART. 48.] THE ELLIPSE. 97 

which shews 1. that if/p is a unit vector; 2. that the equation, of 
the ellipse may be expressed in the form of the equation of a 
circle, the vector which represents the radius being itself of vari- 
able length, deformed by the function \j/. 

Lastly, Sa<j>(3 = 

gives Scupp = Sijnolrp = ; 

therefore if/a, \f/(3 are vectors at right angles to one another. 

48t To find the locus of the middle points of parallel chords. 

Let all the chords be parallel to the vector (3 ; TT the vector 
to the middle point of one of them whose vector length is 2x(3 ; 
then 

TT + xfi, ir xft 
are vectors to points in the ellipse ; 



multiplying out, observing that (44. 1), 

<f> ( 
we get by subtracting, 



= 0, 
or, (Art. 44. 3), 

2Sir<t>P = ; 
.-. Sv<l>p = 0, 
i. e. the locus required is a straight line perpendicular to </>/?. 

Now (f>fi is the vector perpendicular to the tangent at the 
extremity of the diameter ft (Art. 45. Cor. 1). 

Therefore the locus of the middle points of parallel chords is 
a diameter parallel to the tangent at the extremity of the diameter 
to which the chords are parallel. 

COB. If a be the diameter which bisects all chords parallel 
to (3 - } since 

Sa<j>P = 0, 
T. Q. 7 



98 QUATERNIONS. [CHAP. VI. 

we have (Art. 44. 3), 

Sft<j>a = 0, 

which is the equation to the straight line that bisects all chords 
parallel to a. Moreover ft is parallel to the tangent at the ex- 
tremity of a, for it is perpendicular to the normal </>a. 

Hence the properties of a with respect to ft are convertible 
with those of ft with respect to a : and the diameters which 
satisfy the equation 

Sa<t>ft = 0, 

are said to be conjugate to one another. 

49. O ur object being simply to illustrate the process, we shall 
set down in this Article a few of the properties of conjugate 
diameters without attempting to classify or complete them. 

1. If CP, CD are the conjugate semi- diameters a, ft - } and 
if DC be produced to meet the ellipse again in E, and PD, PE 
be joined ; vector DP = a ft, vector EP = a + ft. 

Now 



= Sa<t>a-Sft<}>ft-Sa<j>ft+Sft<l>a(U. 1) 

= o, 

because Safa, Sft^ft, each equals 1. 

Therefore a + ft, a ft are parallel to conjugate diameters. 
(Art. 48. Cor.) 

This is the property of /Supplemental Chords. 

2. Let two tangents meet in T, CI'=-n; and let the chord 
of contact be parallel to ft. If for the present purpose we denote 
CN by a, we have 



(a + a, ft) = 1, 
for the two points of contact. 



ART. 49.] THE ELLIPSE. 99 

Subtracting and applying (44. 1), 

&*<}>($ = : 
hence TT and ft Le. CT, QR are conjugate. 

3. The equation of the chord of contact is S<T^TT = 1. 

For Spfar = 1 (45. Cor. 2) is satisfied by the values of p at 
Q and at B, and since Sp<f>ir = 1 or S<r<}*ie = 1 is the equation 




of a straight line, ir being a constant vector (32. 3) it is the 
line QR. 

4. If QR pass through a fixed point JZ } the locus of T is 
a straight line. 

Let <r be the vector to the point E, then 

Sa-tfrir = 1 ; 
.'. /S f 7r^r = l, 

or the locus of T is a straight line perpendicular to <fxr, i.e. 
parallel to the tangent at the point where CE meets the ellipse. 
(45. Cor. 1.) 

The converse is of course true. 

5. Let us now take 

CP = a, C = p, CN=xa, NQ = y p, CT=za' } 

72 



100 QUATERNIONS. [CHAP. VI. 

then the equation of the tangent becomes 
Sza<j> (xa + yP) = I ; 
i.e. xzSafya = 1 ; 
.: xz = 1, 
or xa.za.-a* ; 
geometrically CN.CT= CP S . 

6. The equation of the ellipse gives 

S(xa + yp) <f> (xa + yp) = 1, 
or x*Sa<}>a 

i.e. 

or, since CN is xa, CP-a, &c., 



\CPJ + \CD. 
the equation of the ellipse referred to conjugate diameters. 

7. a = if/~ l ifra = - (aiSi\f/a + bjSjij/a) 

. '. Vap db Vij (Siij/aSj\}/P Si\ 

If now we call k the unit vector perpendicular to the plane 
of the ellipse, we get 

Vij = k. 

And, observing that ij/a, \f/p are unit vectors at right angles ; 
if the angle between i and tya be 0, that between i and \j/(3 
will be 

- + 6, &c. &c., 

we shall have (21. 3) 

Sifya = COS 6, 

= sin0, 
tya = sin 0, 
= cos 0. 
ida = cos 8 6 + sin* = 1. 



ART. 50.] THE ELLIPSE. 101 

Consequently Fa/3 = able ; 



or all parallelograms circumscribing an ellipse are equal. 

50. EXAMPLES. 

Ex. 1. To find the length of the perpendicular from the centre 
on the tangent. 

Let CY the perpendicular, which (Art. 45. Cor. 1) is a vector 
along </>p, be x<f>p ; then since T is a point in the tangent, 

1 gives /Sx<f>p<f>p = 1, 
or x (<j>p) a = 1 ; 



and 



(46). 



Ex. 2. The product of the perpendiculars from the foci on tJie 
tangent is equal to the square of the semi-axis minor. 

We have SY the vector perpendicular = x<f>p, and as Y is a 
point in the tangent, and 



x (<p) 2 = 1 Saxftp, 



9P 

Similarly, E Z=T l -^-', 

<PP 



102 QUATERNIONS. [CHAP. VI. 

Now (43. 2) V - - S 2 ap -a 4 (I- e 2 ), 
a?p + aSap 



4 Cf% 

r - /3 ap 



Ex. 3. ^Ae perpendicular from the focus on the tangent in- 
tersects the tangent in the circumference of the circle described about 
the axis major. 

Retaining the notation of the last example, we have 
CY=a + x<j>p 

$p(l- Sa<f>p) 

~~ 

2 



., 



= aV a 2 (1 e z ) (last example) 



and the line CY=a. 



Ex. 4. To ^c? i/ie locus of T when the perpendicular from 
the centre on the chord of contact is constant. 

If CT be TT, the equation of QR, the chord of contact, is 
7r=l (Art. 49. 3), 



and the perpendicular (Ex. 1) is T ; 



ART. 50.] THE ELLIPSE. 103 

.-, (**)= -c", 

Or /S<f>7T . <f>TT = C 2 , 

or &r^r = -c* (Art. 44. 3); 






x 2 y* 

r + l? = ^ 

an ellipse. 

Ex. 5. FQ, TR are two tangents to an ellipse, and CQ', CR' 
are drawn to the, ellipse parallel respectively to TQ, TR ; prove 
that Q'R' is parallel to QR. 

Let CQ=p, CR = P ', CT=a, 

then Sp<f>a = 1, 



Now since CQ' ,is parallel to TQ, 

CQ'=xTQ = x(p- 
Similarly CR' = y (p - a), 



and 

gives y?S (p a) <f> (p a) = 1, 

i.e. x*(Sa<j>a-l) = I, 

and y 2 (Sa^a - 1) = 1; 

. . y = x, 



and 

= xQR; 

hence Q'R' is parallel to QR. 
COR. Q'R 2 : QR 3 :: x 2 : 1 

:: 1 : Safa- 



where aj, y are the co-ordinates of T. 



104 QUATERNIONS. [CHAP. VI. 

Ex. 6. If a parallelogram be inscribed in an ellipse, its sides 
are parallel to conjugate diameters. 

Let PQRS be the parallelogram. 



then CQ = p + a, CR = p+a; 

.'. Sp<f>p=l, 



wherefore %Sp<$>a + Sa<^a = 0. 

Similarly 2Sp'<j>a + Sa<j>a = ; 

.'. S(p' p) <a = 0, by subtraction, 

or Sp<f>a = 0, 
and (48. Cor.) /?, a are parallel to conjugate diameters. 

ADDITIONAL EXAMPLES TO CHAP. VI. 

1. Shew that the locus of the points of bisection of chords to 
an ellipse, all of which pass through a given point, is an ellipse. 

2. The locus of the middle points of all straight lines of con- 
stant length terminated by two fixed straight lines, is an ellipse 
whose centre bisects the shortest distance between the fixed lines; 
and whose axes are equally inclined to them. 

3. If chords to an ellipse intersect one another in a given 
point, the rectangles by their segments are to one another as the 
squares of semi-diameters parallel to them. 

4. If PGP', BCD' are conjugate diameters, then PD, PD' 
are proportional to the diameters parallel to them. 

5. If Q be a point in the focal distance SP of an ellipse, such 
that SQ is to SP in a constant ratio, the locus of Q is a similar 
ellipse. 



EX. 6.] THE ELLIPSE. 105 

6. Diameters which coincide -with the diagonals of the paral- 
lelogram on the axes are equal and conjugate. 

7. Also diameters which coincide with the diagonals of any 
parallelogram formed by tangents at the extremities of conjugate 
diameters are conjugate. 

8. The angular points of these parallelograms lie on an ellipse 
similar to the given ellipse and of twice its area. 

9. If from the extremities of the axes of an ellipse four pa- 
rallel lines be drawn, the points in which they cut the curve are 
the extremities of conjugate diameters. 

10. If from the extremity of each of two semi-diameters 
ordinates be drawn to the other, the two triangles so formed will 
be equal in area. 

11. Also if tangents be drawn from the extremity of each 
to meet the other produced, the two triangles so formed will be 
equal in area. 

12. If on the semi-axes a parallelogram be described, and 
about it an ellipse similar and similarly situated to the given 
ellipse be constructed, any chord PQR of the larger ellipse, drawn 
from the further extremity of the diameter CD of the smaller 
ellipse, is bisected by the smaller ellipse at Q. 

13. If TP, TQ be tangents to an ellipse, and PCF be the 
diameter through P, then PQ is parallel to CT. 



CHAPTER VII. 

THE PARABOLA AND HYPERBOLA. 



51. As already stated, most of the properties of the hyperbola 
are the same as the corresponding properties of the ellipse, and 
proved by the same process, e being greater than 1. There are, 
however, some properties both of it and of the parabola which 
may be conveniently developed by a process more analogous to 
that of the Cartesian geometry. This process we shall develope 
presently. In the meantime we proceed to give a brief outline 
of the application to the parabola of the method employed in 
the preceding Chapter for the ellipse. 

52. If S be the focus of a 
parabola, DQ the directrix, we 
have SP = PQ, SA=AD = a. 

If SP = p, SD = a, we have 
(Ex. 5, Art. 35) 

a a p a = (a 3 -SapY (1). 

p a" 1 /Sap 



If <>=' 



(2), 



to which the properties of cj>p in 
Art. 44 evidently apply, 
the equation becomes 

Sp (<f>p + 2a- J ) = 1 




If p r be another point in the parabola, p' p = /?, the 
which /3 approaches is a vector along the tangent ; so 
xf} = ir-p, TT is the vector to a point in the tangent ; this 



..(3). 

limit to 
that if 
gives 



ART. 52.] THE PARABOLA AND HYPERBOLA. 107 

hence the equation of the tangent becomes 

Sir(<l>p+a- l )+Sa- l p=l ................... (5). 

From (2) it is evident that 



so that <f>p is a vector perpendicular to the axis. 
From the same equation 



a 



From (4) the normal vector is 

tp+a- 1 ............................ (8); 

therefore the equation of the normal is 

<r = p + x (<t>p + a" 1 ) ....................... (9). 

Equation (2) when exhibited as 

a 2 (f)p = p a~ l /Sap, 

reads by (6), 'vector along NP = SP - vector along AN', which 
requires that 

a*<j>p ............................ (10), 



i.e. =cuSa-*p ......................... (11). 

For the subtangent AT, put xa for TT in (5), and there results 
by (6) 

X + Sa l p = ~i, 

whence \ x ~ 9J a = 9 a "~ a ^ a ~V > 

i. e. vector A T = - vector AN (by 11); 



108 QUATERNIONS. [CHAP. VII. 

and ST=xa gives 

S2" = (a-aSar l p) a 
_(a'-Sap) s 



.-. line ST=SP, 

whence also the tangent bisects the angle SPQ ; and SQ is per- 
pendicular to and bisected by the tangent. 

From (8) y ($p + a~ l ) = PG 

= PN+ NG 

= - a a <f>p + zo. (by 10) ; 

.-. y = -o*, y = za s , 

i. 

za = a ) 
Le. NG = -SD, 
or linQNG = SD, 
whence the subnormal is constant. 
And vector GP --y (<f>p + a~ l ) - a? (<j>p + a" ') ; 
.-. vector SQ = SD+DQ 



and SQGP is a rhombus. 
Lastly, 



= a + a.*(j)p 



or (10) A Y is parallel to, and equal to half of NP. 



ART. 54.] THE PARABOLA AND HYPERBOLA. 109 

53. If n ow we substitute Cartesian co-ordinates, making 

p = xi + yj, a = -2ai; 
we shall have 

C* 1 *^ 

^ a p =~2a' 
a~ 1 Sap = xi, 

^ = ~^ ; 
and equation (3) becomes 

J^_*_l 
4a 8 a 

or y* = a(a + x) 

= 4ax' if x' = AN. 



The locus of the middle points of parallel chords is thus 
found. 

Let the chords be parallel to (3, TT the vector of the middle 
point of one of the chords, 

then 

and X 

which, since the term involving x must disappear, gives 



a straight line perpendicular to </?, i. e. (6) parallel to the axis. 
This equation may be written 

tf (<*+ a" 1 ) = 0, 

which shews (8) that the chords are perpendicular to the normal 
vector at the point where P = TT, i.e. at the point where the 
locus of the chords meets the curve : in other words, the chords 
are parallel to the tangent at the extremity of the diameter which 
bisects them. 

54. EXAMPLES. 

Ex. 1. If two chords be drawn always parallel to given lines, 
and cut one another at points either within or urithout the parabola, 



110 QUATERNIONS. [CHAP. VII. 

the ratio of the rectangles of their segments is always the same 
whatever be their point of section. 

Let POp, QOq be the chords drawn through 0, and always 
parallel respectively to /3 and y, which we will suppose to be 
unit vectors. 

Let 8 be the vector to 0, 
then p-S + x/3 

gives from equation (3) 



the product of the two values of x being 



a constant ratio whatever be 0. 

Con. Let 6, & be the angles in which P and y cut the axis ; 
then since /3, y are unit vectors, if p be a vector to the parabola, 
drawn from S parallel to POp, which we may now call SP ; 

P = n(3, <fr> = ^(w) = n^8(44. 2), 
will ive 



. NP 

in which case <pp is j- ; 

a 



: Sy<f>y :: sintf-^ : sintf'-^- :: sin 2 : sin 2 6'; 

and, OP . Op : OQ . Oq :: . a/i : . a/v . 

sin sin o 

Ex. 2. jFwc^ f/ie locus of the point which divides a system of 
parallel chords into segments whose product is constant. 



AET. 54.] THE PAEABOLA AND HYPERBOLA. Ill 

By the last example, the equation of the locus is 



a parabola similar to the given parabola. 

Ex. 3. The perpendicular from A on tJte tangent, and the line 
PQ are produced to meet in R : find the locus of R. 

By Art. 52. 8, AR = x (<f> P + a' 1 ), 
and PR = ya ; 

~ 



Operate by S<j)p, 
and x (<p) 2 = Sp<j>p 

(52.7); 



and TT = <T + a 2 (<j>p + a" 1 ) 



O 

= -H- + a*^P is the equation required ; 



(O V 
TT ^- j a 0, it is that of a straight line perpendi- 

cular to the axis, at the distance 3a from 8. 



Ex. 4. Jb ^/md tf/^e focws of the intersection with t/te tangent 
of the perpendicular on it from the vertex. 

If TT be the vector perpendicular on the tangent from A, 
we have by (52. 8) 

TT = x (0p + a' 1 ) .......................... (1), 

and the equation of the tangent gives, putting TT + ^ in place 

a 

of TT in (52. 5), and multiplying by 2, - 

2/ffir^p + 2/S r a-'ir + a^o^p = 1 ................. (2), 

we have also 

Sp (<j> P + 2a- J ) = 1 ..... ................ (3). 



112 QUATERNIONS. [CHAP. VII. 

From these three equations we have to eliminate x and p. 

Equation (1) gives 

SO.TT - x, 
which gives x, 
and . Str(f)p = x (<f>p) 2 , 

which substituted in (2) gives 



Also, substituting (52. 7) a* (<j>p) 3 for Sptfrp, equation (3) 
gives 



therefore by subtraction 

(2x - a 2 

i. e. (2Sa.Tr - a 2 ) (<f>p)* + 2Sa- l ir = 0, 
which from (1) becomes, multiplying by S*air, 

(2Sair - a) 2 (n - aT l Sa.Tr} 2 + 2S 2 a7nS f cr 1 7r = 0. 
This equation at once reduces to 

27r e #x7r - TT-V + S*cnr = 0, 

an equation which, when 4a is written in place of a, becomes 
identical with that obtained in Art. 37, Ex. 8. 

The locus is therefore a cissoid, the diameter of the generating 
circle being AD. 

55. It will probably have suggested itself to the reader, that 
there exists a large class of problems to which the processes we 
have illustrated are scarcely if at all applicable. Hence there 
may have arisen a contrast between the Cartesian Geometry and 
Quaternions unfavourable to the latter. To remove this un- 
favourable impression, all that is required in a reader familiar with 
the older Geometry is a little experience in combining the logic 
of the new analysis with the forms of the old. He will then see 
how simple and direct are the arguments which he can bring 
to bear on any individual problem, and consequently how little 
the memory is taxed. 



ART. 55.] THE PARABOLA AND HYPERBOLA. 113 

We propose in this Article to put the reader in the track 
of employing his old forms in conjunction with quaternion 
reasonings. 

We shall work several examples on the parabola and the 
hyperbola. Having applied quaternions pretty fully to the 
ellipse in what has preceded, we will limit ourselves to a single 
example in this case. 

1. The Parabola. If the unit vector along any diameter of 
the parabola be a, and the unit vector parallel to the tangent at 
its extremity be ft; we may write the equation of the parabola 
under the form 



For the particular case in which the diameter in question is the 
axis, and the tangent at its extremity parallel to the directrix 

where a is A S (Art. 52). 

This is the most convenient form when the focus is referred 
to. 

In other cases a somewhat simpler form may be obtained by 
supposing a, or if necessary both a and /3 of equation (1) to 
be other than unit vectors. 

The equation may then be written under the form 

P = 2* + *P (3). 

To find the equation of the tangent, we have 



T. Q. 



QUATERNIONS. [CHAP. VII. 

Now p p is a vector along the secant; and its limit is a 
rector along the tangent : hence any vector along the tangent 
is a multiple of to. + /? ; and the equation of the tangent may 
be written 



(4). 



EXAMPLES. 

Ex. 1. If AP, AQ be chords drawn at rigid angles to one 
another from A ; PM, Q<& perpendiculars on tJie axis, then the 
latus rectum is a mean proportioned between AM and AN ; or 
between PM and QN. 

If PJf=y, QK=y, 



, 
Now S(AP.AQ)=0(22. 7); 



or yy= 
therefore also aai - 

Ex. 2. If the rectangle of ichich AP, AQ are the fides be 
completed, the further angle witt trace out a parabola similar to 
the given parabola, tfe distance between the tico vertices being equal 
to twice the latus rectum. 



Ex. 3. The circle described on a focal chovd as diameter touch fs 
the directrix; and the circle described on any other chord dots 
not reach tfte directrix. 



ART. 55.] THE PARABOLA AND HYPERBOLA. 115 

Let PQ be any chord, centre 0, 






The equation of the circle with centre 0, radius OP, is 

AQ-AP\* 



/ 
( p - 



2 
or p-S(AP + 

At the points in which this circle meets the directrix 
p = aa -f s/3 ; 



or 



This equation is possible only when 

yy+4a*=0; 
i. e. when the chord is a focal chord. 

I/ J. T/' 

In this case the two values of z are equal, each being ( - ; 

A 

and the directrix is a tangent to the circle. 

Ex. 4. Two parabolas have a common focus and axis ; their 
vertices are turned in opposite directions. A focal chord cuts 
them in PQ, P'Q', so that PP'SQQ' are in order. Prove (1) that 
SP.SP = SQ.SQ'; (2) that SP : SQ' ia a constant ratio; and 
(3) that the tangents at P, f are at right angles to one another. 

The equations of the parabolas are 

V 3 
+ 



the focus being the origin. 

82 



116 QUATERNIONS. [CHAP. VII. 

Now since p, p are in the same straight Hue when the common 
chord is the focal chord, we have 

p'=pp; 



y'=py, 

' (yy' - 4oa') (ay + ay') = 0. 

Taking the former factor, we must have y, y' on the same 
side of the axis with a constant product; therefore 



The second factor gives SP : SQ' a constant ratio a : a'. 

Lastly, by Equation (4), the tangent vectors at P and P' are 
parallel to 



therefore the tangents are at right angles to one another. 

Ex. 5. If a triangle be inscribed in a parabola, the three 
points in which ilie sides are met by the tangents at the angles lie 
in a straight line. 

Let OPQ be the triangle. 
Take as the origin, then 




t* 
r=2<>- 

t r * 
' ' ' 



ART. 55.] THE PARABOLA AND HYPERBOLA. 117 

are the vectors OP, OQ, and the equations of the tangents at P 
and Q. 

If QO meet in A the tangent at P, 
t 2 





9 y> 

t + x- t'y, 

t 3 



and 



Similarly if the tangent at Q meets PO in B, 



If the tangent at meets PQ in (7, 
OC=OP + z(PQ) 

= OP + z(OQ-OP) 

t 3 (t" t 2 

= -o + $ + *| 2 a + 

But OC = v(3; 



2 T ~2~ 
t + z(^ t) v, 
ft' 



QUATERNIONS. [CHAP. VII. 



Now 



2t-t' It'-t f-t" 
and also --- j- -g- 0; 

therefore (Art. 13) A, B, C are in a straight line. 

2. The ellipse. If a, ft are unit vectors along the axes, the 
equation of the ellipse may be written 



b* 
where y* = -5 (a 2 - a; 8 ) = m (a 2 - # 2 ) ; 

Gb 

and the equation of the tangent will be readily seen to be 

ir = xa + y(3 + X (ya - mxfi). 
A single example will suffice. 

Ex. If tangents be drawn at three points P, Q, R of an 
ellipse intersecting in R', Q', P, prove tJiat, 

PR'. QF. RQ' - PQ'. QR'. RP. 

If x, y; x', y ; x", y" are respectively the co-ordinates of 
P, Q, R', we shall have 

CR' xa + y($ + X (ya. - mxft) 
- x'a + y'fi + X ' (y'a - mx'fi) ; 



y mXx = y' mX'x' y 
. \ mZ (x'y - y'x) = mx' a + y' 2 - mxx' - yy' 

= b a mxx' yy' . 
Hence mX ' (xy - x'y) -b a - mxx' - yy' 

= -mX(xy'-x'y); 
.-. X=-X', 

Y = -Y' for<?', 
Z = -Z' for/", 



and 



ART. 55.] THE PARABOLA AND HYPERBOLA. 



119 



Now 



X PR' . 
= 



hence the proposition. 

3. The hyperbola. If a, 8 are unit vectors parallel to the 
asymptotes CX, CY, the equation of the hyperbola may be written 



since 



= xa + - B. 
x 

,= a ^-=G. 



If a, /3 be not both units we may write the equation under 
the simpler foi*m 

P = a + .............................. (1). 

To find the equatioa of the tangent, we have as usual a vector 
parallel to the secant 



and a vector parallel to the tangent will be 




120 QUATERNIONS. [CHAP. VII. 

Hence the equation of the tangent is 

TT = to. + - + x ( to. - '-, 
t 

COR. It is evident that 



are conjugate semi-diameters. 

EXAMPLES. 

Ex. 1. One diagonal of a parallelogram tcJtose sides are tJie 
co-ordinates being the radius vector, the other diagonal is parallel to 
the tangent. 

We have CN = ta, tfQ = % , 

t 

7> 

and the other diagonal is 

which, equation (2), is parallel to the tangent at Q. 

Ex. 2. Any diameter CP bisects all the chords which are 
parallel to the tangent at P. 

Let CP be to. + - , 

t 

then the tangent at P is parallel to 

.-f; 



But as Q is a point in the hyperbola, this equation must have 
the form 



ART. 55.] THE PARAIOLA AND HYPERBOLA. 121 



and X*-Y a =l, 

an equation which gives two equal values of Y with opposite 
signs, for every value of X. 

Hence all chords are bisected. 
COR. X'-Y a = lia 

f2KJ/fi? 

\VP) \CD 

CD being ta-@ = PO. 

t 

This is the ordinary equation of the hyperbola referred to 
conjugate diameters. 

Ex. 3. If TQ, T'Q' be two tangents to tJte hyperbola intersect- 
ing in R and terminated at T, T', Q, Q' by the asymptotes; then 
(1) TQ' is parallel to T'Q; (2) area of triangle TRT' = area of 
triangle QRQ', and (3) CR bisects TQ' and T'Q. 

The equation of the tangent 



gives 

fiW Of. 
I/ J. Ml 

(the coefficient of /3 being 0), 

t 
CT = 2t' 






* v 

therefore Q'T is parallel to QT'. 



122 QUATERNIONS. [CHAP. VII. 

Again, CR = CQ+QR = CQ + 



IB f B\ 

Also CR = ~ + x'2(at'- p ,}; 

t \ t 

.: xt = x't', 
1 x 1 tf 



t' 



~t+t" 

,_ t 
~ t + 1'' 

and xx' = ( 1 - a;) (1 a;'), 

and the triangles TRT', QRQ' are equal 
Lastly, C 



- 
t t+t\ t 



or CR is in the direction of the diagonal of the parallelogram of 
which the sides are CT, CQ' ; and therefore CR bisects TQ' 
and T'Q. 

Ex. 4. If through Q, P, Q' parallels be drawn to CX meeting 
CY in E, F, G ; CE, CF, CG are in continued proportion. 



t 



= GV+VQ 



ART. 55.] THE PARABOLA AND HYPERBOLA. 123 



, 



CF-* 
I 1 



and CE.CG=CF 2 ; 

because X*-Y'=l (Ex. 2). 

Ex. 5. If a chord of a hyperbola, be one diagonal of a 
parallelogram whose sides are parallel to the asymptotes, the other 
diagonal passes through the centre. 

Let the chord be PQ ; p, p the vectors to P and Q ; then 



t 

Now when one diagonal of a parallelogram is ma + n(3, the 
other will be ma n(3. 

Therefore in the case before us, the other diagonal is 



! -J) 



And it is therefore in the same straight line with the line 
which joins the centre of the hyperbola with the middle point 
of PQ ; whence the truth of the proposition. 



124 QUATERNIONS. [CHAP. VII. 

Ex. 6. If two tangents to a hyperbola at the extremities 
Qi Q' of <*> diameter, meet a tangent at P in the points T, T'; 
and if CD, CD' are the semi-diameters conjugate to CP, CQ ; 
tJten (1) PT : QT :: PT' : Q'T' :: CD : CD'- 

and (2) PT.PT' = CD\ 

If t, t', t', correspond to P, Q, Q', then 



; ~^) 



gives t + xt = t' + x't', 

1 _x 1 _ni_ 
~i~~i~t'~7' 

t' -t 

x= 7^t = ~ x ' 

Similarly CT' = at + & + y fat - 



gves 

I y I y' 

__._ __ + 

it t' t" 

whence -, 

v r 

Now x : y :: x : y' 

gives PT : QT :: PT' : QT' 

:: CD : CD'. 

And a:?/ = 1 

gives PT.PT'=CD\ 

COR. x'y'=l, 

gives QT.Q'T'=CD\ 



ART. 55.] THE PARABOLA AND HYPERBOLA. 125 

Ex. 7. Straight lines move so that the triangular area which 
they cut off from two given straight lines which meet one another 
is constant: to find tlie locus of their ultimate intersections. 

Let OAA', ORE' be the fixed lines, AB, A'B'two of the moving 
lines with the condition that 

OA.OB = OA'.OB\ 
If a, /3 be unit vectors along OA, OB, 

OA = ta, OB = u p-, OA' = t' a , OK = u'p, 
the point of intersection of AB, AB' gives 
p = to. + x (u(3 to) 
= t'a + x' (u'(3 - t'a), 
.'. XU = x'u, 

and t (1 - x) = t' (1 - x') 



Now tu = t'u' = c because the triangle has a constant area; 

. *. x = - --, = - ultimately; 

t + 1 '-i 



the equation of a hyperbola. 

ADDITIONAL EXAMPLES TO CHAP. VII. 

1 . In the parabola SY 2 = SP.SA. 

2. If the tangent to a parabola cut the directrix in fi, SH is 
perpendicular to SP. 

3. A circle has its centre at the vertex A of a parabola whose 
focus is S, and the diameter of the circle is 3AS. Prove that the 
common chord bisects AS. 

4. The tangent at any point of a parabola meets the directrix 
and latus rectum in two points equally distant from the focus. 



126 QUATERNIONS. [CHAP. TIL 

5. The circle described on SP as diameter is touched by the 
tangent at the vertex. 

6. Parabolas have their axes parallel and all pass through 
two given points. Prove that their foci lie in a conic section. 

7. Two parabolas have a common directrix. Prove that 
their common chord bisects at right angles the line joining their 
foci. 

8. The portion of any tangent to the parabola between tan- 
gents which meet in the directrix subtends a right angle at the 
focus. 

9. If from the point of contact of a tangent to a parabola 
a chord be drawn, and another line be drawn parallel to the axis 
meeting the chord, tangent and curve ; this line will be divided 
by them in the same ratio as it divides the chord. 

10. The middle points of focal chords describe a parabola 
whose latus rectum is half that of the given parabola. 

11. PSQ is a focal chord of a parabola: PA, QA meet the 
directrix in y, z. Prove that Pz, Qy are parallel to the axis. 

12. The tangent at D to the conjugate hyperbola is parallel 
toCP. 

13. The portion of the tangent to a hyperbola which is in- 
tercepted by the asymptotes is bisected at the point of contact. 

14. The locus of a point which divides in a given ratio lines 
which cut off equal areas from the space enclosed by two given 
straight lines is a hyperbola of which these lines are the asymp- 
totes. 

15. The tangent to a hyperbola at P meets an asymptote 
in T, and TQ is drawn to the curve parallel to the other asymp- 
tote. PQ produced both ways meets the asymptotes in R, R : 
RR is trisected in P, Q. 



ART. 55.] THE PARABOLA AND HYPERBOLA. 127 

16. From any point JK of an asymptote, UN, EM &TQ drawn 
parallel to conjugate diameters intersecting the hyperbola and its 
conjugate in P and D. Prove that CP and CD are conjugate. 

17. The intercepts on any straight line between the hyper- 
bola and its asymptotes are equal. 

18. If QQ' meet the asymptotes in R, r, 



19. If the tangent at any point meet the asymptotes in X 
and I 7 , the area of the triangle XCY is constant. 



CHAPTER VIII. 

CENTRAL SURFACES OF THE SECOND ORDER, PARTICULARLY 
THE ELLIPSOID AND CONE. 

56. The Ellipsoid. In discussing central surfaces of the 
second order, we shall speak as if our results were limited to the 
ellipsoid. That such limitation is not, in most cases, necessarily 
imposed on us, will be apparent to any one who has a slender 
acquaintance with ordinary Analytical Geometry. We adopt it 
in order that our language may have more precision, and that, in 
some instances, our analysis may have greater simplicity. If the 
centre be made the origin it is clear that the scalar equation can 
contain no such term as ASap, for the definition of a central sur- 
face requires that the equation shall be satisfied both by + p and 
by -p. 

If we turn to the equation of the ellipse (Art. 43), we shall 
see at once that the equation of the ellipsoid must have the form 
ap* + bS'ap + 2cSapSp P + ... = 1. 

Now if, as in the Article referred to, we put 

<f>p = ap + baSap + c (aS(3p + (3Sap) + ... 
we shall have 

Sp<l>p = ap* + bS*ap + 2cSapS(3p + ... 

-li 

the equation required. 

It will be seen that, as in Arts. 32, 33, one form of the equa- 
tion of the straight line was found to coincide exactly with the 
equation of a plane, so a form of the equation of the ellipse 
coincides exactly with the equation of the ellipsoid. 



ART. 58.] CENTRAL SURFACES OF THE SECOXD ORDER. 129 

It is evident that the three properties of </a given in Art. 44 
are true of </>p in its present form. 

57. To find the equation of the tangent plane. 

Let a secant plane pass through the point whose vector is p; 
and let p be the vector to any point of section. 

Put p - p + (3, where /3 is a vector along the secant plane ; 
then Sp'tp = S(p 

Hence, observing that (44) 



and 

we have Sp'fo' = Sp<f>p + '2S(3(j>p + 



Now (45), as the secant plane approaches the tangent plane, 
the sum of these two expressions approaches in value to the first 
alone : that is, for the tangent plane, S{3<f>p = 0, where /? is a vector 
along that plane. 

If TT be the vector to a point in the tangent plane, 



.'. S (ir p) </>p = xS{3<f>p 
= 0, 

and Sir<j>p = Sp(f>p 

- i 
is the equation of the tangent plane. 

COR. <f>p is a vector perpendicular to the tangent plane at the 
extremity of the vector p. 

58. If OF be perpendicular from the centre on the tangent 
plane; then, since <f>p is a vector perpendicular to that plane, 
OY- x<f>p and Sx (<f>p)* - 1, giving 

. or-rwrt-rl. \: 

Sir W. Hamilton terms <$>p the vector of proximity. [In fact 
vector OF 
T. Q. 



130 QUATERNIONS. [CHAP. VIII. 

59. If tangent planes all pass thnnigh a fixed point, the 
curve of contact is a plane curve. 

Let T be the fixed point ; vector a ; p the vector to a point of 
contact. 

Then (Art. 57) Sa<f>p = 1 ; 

i.e. Sp^a=l (44. 3), 
which is the equation in p of a plane perpendicular to </>a. 

Now </>a is the normal vector of the point where OT cuts the 
ellipsoid ; 

.. the curve of contact lies in a plane parallel to the tangent 
plane at the extremity of the diameter drawn to the given point. 

The plane of contact is called the polar plane to the point. 

60. Tangent planes are all parallel to a given straight line, 
to find the curve of contact. 

Let a be a vector parallel to the given line ; then 

TT p + xa 
is a point in the tangent plane ; 

.'. S(p + xa) (f>p= 1 ; 

and Sa<f>p = 0, 

or /Sp(f>a = 0, 

the equation of a plane through the origin perpendicular to <a : 
that is, the curve of contact lies in a plane through the centre 
parallel to the tangent plane at the extremity of the diameter 
which is parallel to the given line. 

61. To find the locus of the middle points of parallel chords. 
Let each of the chords be parallel to a, it the vector to the 

middle point of one of them j then TT + xa, tr xa are points in 
the ellipsoid. 

From the first, 

S(ir + xa) <f>(ir + xa) = l (Art. 56) j 
i. e. Sir<f>ir + 2xSrr< 



ART. 61.] CENTRAL SURFACES OF THE SECOND ORDER. 131 
From the second, 



.'. subtracting, S-n-(f>a. = Q (1), 

i. e. the locus is a plane through the centre perpendicular to <a, 
or parallel to the tangent plane at the extremity A of the 
diameter which is drawn parallel to a. 

If we call this the plane BOC, B and C being any points in 
which it cuts the ellipsoid ; and if OB = (3, 00= y, we shall have 



and therefore Sa<f>(3 = 0, 

or a satisfies the equation >&r</3 - 
of the plane which bisects all chords parallel to OB (Equation 1). 

Let AOC be this plane which bisects all chords parallel to OB. 

Then, since 00 or y is a vector in it, 



But we have already proved that 

iSytfta = 0, i. e. Sa.(f>y = 0, 
because y is in the plane BOC ; 

.-. by equation (1) a, (3 both satisfy the equation of the plane 
Sir<j>y = 0, which is the plane bisecting all chords parallel to y ; 
that plane is therefore the plane AOB: we are thus presented 
with three lines OA, OB, OC such that all chords parallel to any 
one of them are bisected by the diametral plane which passes 
through the other two. 

"We may term these lines conjugate semi-diameters, and the 
corresponding diametral planes conjugate diametral planes. 

It is evident that the number of conjugate diameters is 
unlimited. 

COB. We have the following equations : 



(2). 

92 



132 QUATERNIONS. [CHAP. VIII. 

They shew that y is perpendicular to both <a and <f>ft, and is 
therefore a vector perpendicular to their plane ; hence, as in 34. 4, 
y = X V(}>a<j>ft. 

In the same way, since <y is perpendicular to both a and ft, 
we have 



or, neglecting tensors, we have the following vector equalities : 
y = V<j>a<}>/3, ft = F<a<y, a = V<j>ft<f>y, 
<y = Fa/3,. ^ft= Fay, <a = Vfty (3). 
Note also 



upon which Hamilton founded his solution of linear equations. 

62. If as in Art. 47 we write \}/\(/p for <f>p, ij/p being still a 
vector, the equation of the ellipsoid assumes the form 



i. e. (44) Sif/pif/p - I 

(^)-=-r(^)'=-i ............ (i), 

which, if we put <r = ij/p, becomes To- ~ 1, the equation of a sphere. 

Hence the ellipsoid can be changed into the sphere and vice 
versd, by a linear deformation of each vector, the operator being 
the function i^ or its inverse. 

The equations 



now become ScuJ/ 2 ft = 0, 

i.e. Siffauj/ft = 0, &c., &c .................... (2). 

(1) and (2) shew that \[/a, ij/ft, \j/y are unit vectors at right angles 
to one another. 

If we term the sphere To- = 1 the unit-sphere, we may 
enunciate this result by saying that the vectors of the unit-sphere 
which correspond to semi- conjugate diameters form a rectangular 
system. 



ART. 63.] CENTRAL SURFACES OF THE SECOND ORDER. 133 

63. Let us now take i, j, k unit vectors along the principal 
axes of x, y, z ; then we shall have 



(1), 
. '. Sip = x, &c. 

so that for the sake of transformations in which it is desirable 
that the form of p should be retained, we may write 

p = -(iSip+jSjp + kSkp) .................. (2); 

and as <{>p is a linear and vector function of p, its vector portions 
along the principal axes will be multiples of 

iSip, jSjp, kSkp ; 
we may therefore write 



the form a 2 having been assumed in order to make the equation 

Sp<f>p = 1 

coincide with the Cartesian equation 

x 3 if z 3 
__ i_ y __ L _ i 

a 3 + b* <r 



(4), 

we require to take \j/p so that performing the operation if/ twice 
on p shall give the same result (with a - sign) as performing the 
operation < once. 

Now a comparison of equations (2) and (3) will shew that 
the latter operation introduces -5 &c. into p ; it is evident 



therefore that the former operation (^) is to introduce - &c. or 



\ a 



134 QUATERNIONS. [CHAP. VIII. 

It may perhaps be worth while to verify this result. We have 

fJStyp jSjtp kSfyp\ 
wilrp = I -- 1 -- - -- 1 -- 

\ a b c J 



a\ a b c / 

.i'Sip 

= t -/+... 
a 

fiSip jSjp kSkp\ 

~~ + ~~^ ~ 



/iSip jSjp JcSkp\ 

"" " 1 ~' 



(7), 
because <f><j>~ l p produces p. 

\j/~*p = - (aiSip + bjSjp + ckSkp) ................... (8 ), 

(9). 



It is evident that the properties of Art. 44 apply to all these 
functions. 

64. EXAMPLES. 

Ex. 1. Find the point on an ellipsoid, the tangent plane at 
which cuts off equal portions from the axes. 

Let x, y, z be the co-ordinates of the point, p the portion cut 
off, then 

p = xi + yj + zk. 

Now pi, pj, pk are points on the tangent plane ; 

.'. Spi<f>p= I, 
which gives 



ART. 64.] CENTRAL SURFACES OF THE SECOND ORDER. 135 

or --= = 1. 

a 

Similarly -^ = 1, 

? = ! 

x y z 1 1 

a 2 b 2 c*~ p~ Jcf^t-lf + tf ' 

Ex. 2. To find the perpendicular from the centre of the 
ellipsoid on a tangent plane. 

1\ 8 



OY a =(T-=-\ ; (Art. 58) 
\ 9P/ 



- ^+ f! + ^ (Art. 63, 1. 3). 

Ex. 3. To ^/mcJ the locus of the points of contact of tangent 
planes which make a given angle with tlie axis ofz. 

"We have 



Z* ,/X 1 I/* Z* 
Or -*=P - + Ii + 

c \a* b c 

the equation of a cone whose axis is that of z and guiding curve 
an ellipse whose semi-axes are a 2 , b 3 . 

The intersection of this surface with the ellipsoid is the locus 
required. 

Ex. 4. To find the locus of a point when the perpendicular 
from the centre on its polar plane is of constant length. 

Let TT be the vector to the point, then 

= 1 is the equation of the polar plane (Art. 59), 



and T - is the length of the perpendicular on it (Art. 58) ; 

<f>7T 



13G QUATERNIONS. [CHAP. VIII. 

.'. S(<j>irf = -C*, by the question. 
But since (44) 

if 8 be (f>ir, 



. : /Sir^ir C a is the equation required ; 
hence the Cartesian equation is (63. 6) 

T 2 II 2 Z? 

x >y , z r* 

~i + Ti "* -- i~ ^ 

a b c 

Ex. 5. The sum of the squares of three conjugate semi-dia- 
meters is constant. 

Let a, /3, y be the semi-diameters ; i/^a, i^/3, I/Q/ are rectangular 
unit vectors (Art. 62). 

Now a = - (aiStya + bjSfya + ckSkfya) (63. 9) ; 
. '. (Ta) 2 = - a 2 = a 2 (Stya)" + I 2 (Sj^a) 2 + c 2 
a* (StyP) 2 + VfiJtPY + c 2 
a 2 (Sty?)' + b 2 (SJty) 2 + c 2 (Styy? : 
adding, and observing that 

yxtfF+(afyfF+(8fytf*i (si. cor.), 

we get 

(To) 2 + (Tpy + (Ty) 2 = a? + b 2 + c 2 , 



Ex. 6. The sum of the squares of the three perpendiculars from 
the centre on three tangent planes at right angles to one another is 
constant. 

We have 

p - ^- l <f>p = a'iSfyp + b 3 jSj<f>p + c 2 JcSk(j>p (63. 7), 
and <t>p = - (iSi^p +jSj<f>p + kSk<f> P ) (63. 2) ; 

.*. Sp<j>p = 1 - a 3 (SUfrp)' + b (Sjfo)' + c 2 (Sk<t>p)' 

{a 2 (SiUfr)* + b 2 (Sj Ufa)* + c 2 (Sk Ufa)*} ; 



ART. 64.] CENTRAL SURFACES OF THE SECOND ORDER. 137 

hence if p, p, p" be three vectors so that <p, (f>p, <pp" are at right 
angles to each other ; that is, so that the tangent planes at their 
extremities are at right angles to one another (57. Cor.), 
1 1 1 



w 

- a 2 {(Sil7<j> P )' + (SiU^p') 2 + ( 



= a ' + b* + c' (31. Cor.). 
But , &c. are the perpendiculars from the centre on the 



tangent planes at p, p', p" (58). Hence the proposition. 

Ex. 7. The siim of the squares of the projections of three con- 
jugate diameters on any of the principal axes is equal to the square 
of that axis. 

Let a, ft, y be conjugate semi-diameters; then, since 
a = - (aiSiij/a + bjSjta + ckSktya) (63. 9), 

Sia. = aStya. 

Similarly, Si(3 = aSiij/fi, 

Siy aSiij/y j 
.-. (Sia) 3 + (Si/3) 3 + (Si 7 ) 2 = a 2 {(Stya)' + (StyP)* + (Sty?)*} 

= a 2 (31. Cor.), 
because ij/a, \f/(3, \f/y are at right angles to one another (62). 

But Sia is the projection of Ta along the axis of x; and 
similarly of the others. Hence the proposition. 

Ex. 8. The sum of tlie reciprocals of the squares of the three 
perpendiculars from the centre on tangent planes at the extremities 
of conjugate diameters is constant. 

Let Oy lt Oy a , Oy a be the perpendiculars. 
J-, = -(<K>* (58) 

(SiaY (SjaY (Ska)' 

~ "~ otf.j 



138 QUATERNIONS. [CHAP. VIII. 

i 



_ 

Oy*~ a 4 6 4 



- . 

4 4 4 



_- 
0y a " a 






= OSW) 2 + (Si/3? + (%)* + &c. 



= -,+ i+-, (Ex.7). 
a o~ c 

Ex. 9. If through a fixed point within an ellipsoid three 
chords be drawn mutually at right angles, the sum of tJie recipro- 
cals of the products of their segments will be constant. 

Let 6 be the vector to the given point ; a, (3, y unit vectors 
parallel to three chords at right angles to each other. 

Then 6 + xa = p gives 



a quadratic equation in x, the product of whose roots is 

-I 



. '. the product of the reciprocals of the segments of the chord is 
1 $a<f>a 1 



-1 ' (To.) 2 ' 
and the sum of the reciprocals of the products of the segments is 



(Sia)' (SjaY (Ska) 2 .. 

Now since SaAa = * ^ + ~r~ + r~ ( 63 - 2 J 3 )> 
2 



r~ 
6 c 

the sum of the reciprocals of the products 






ART. 64.] CENTRAL SURFACES OF THE SECOND ORDER. 139 



- ' 

- I \ct 



- Cor ')- 



COB. If be not constant, but S6<j>0 be so, i. e. if the given 
point be situated on an ellipsoid concentric with and similar to the 
given ellipsoid, the same is true. 

Ex. 10. If the poles lie in a plane parallel to yz, the polar 
planes cut the axis of x always in the same point. 

Let pi be the distance from the origin of the plane in which 
the poles lie, 8 any line in that plane, then ir=pi + 8 is the vector 
to a pole, and 

S P <t>(pi + o) = l (59) 

the equation of the corresponding polar plane. 

At the point where this plane cuts the axis of a?, 

p = xi; 
. . Spxi<f>i + xSi<f>8 = 1 . 

Now 8 is a vector in a plane perpendicular to <j>i, 



and Si<f>i= constant = n suppose ; 

.'. npx= 1, 
which shews that x is constant. 

Ex. 11. A, B and C are three similar and similarly 
situated ellipsoids; A and B are concentric, and C has its centre 
on the surface of B. To shew that the tangent plane to B at this 
point is parallel to the plane of intersection of A and C. 

Let a be the vector to the centre of C. 

= a the equation of A, 



S(p-a)<p(p-a)=c ...... (7. 



140 QUATERNIONS. [CHAP. VIII. 

Now at the intersection of A and C, p is the same for both ; 
therefore the equation of the plane of intersection is to be found 
by subtracting the one from the other. 

It is therefore 2/Sp<f>a = Sat/to. + a-c ; 

and the equation of the tangent plane to B at the centre of C is 
Srr^a b ; 

.: both planes are perpendicular to <a, and are consequently 
parallel. 

Ex. 12. If through a given point chords be drawn to an 
ellipsoid, the intersections of pairs of tangent planes at their ex- 
tremities all lie in a plane parallel to the tangent plane at the 
extremity of the diameter which passes through the point. 

Let a be the vector to the point ; a + xfl, a + xfl, the vectors 
to the points of intersection with the ellipsoid of chords parallel 
to ft ; then 

STr<f> (a 4 a;^) = 1, 



are the equations of the tangent planes at these points. 

At the intersection of these planes w is the same for both ; 
.'. subtracting we get 

Sir<f>{3 = 0, 
STT^CL = 1 . 

The last equation is that of the line of intersection of the tan- 
gent planes; and that line is perpendicular to tf>a, or (57. Cor.) 
parallel to the tangent plane at the extremity of the diameter 
which passes through the given point. 

COR. S-7r<j>(3 shews that the line of intersection correspond- 
ing to any one chord is parallel to the tangent plane at the 
extremity of the diameter which is parallel to that chord. 

Ex. 13. Two similar and similarly situated ellipsoids are cut 
by a series of ellipsoids similar and similarly situated to the two 



ART. 64.] CENTRAL SURFACES OF THE SECOND ORDER. 141 

given ones ; and in such a manner that the planes of intersection 
are at right angles to one another. Skew that the centres of the 
cutting ellipsoids lie on another ellipsoid. 



Let Sp<t> P = l ............................ (1), 

S(p-a)4>(p-a) = C ................... (2), 

be the given ellipsoids; 

S(p-ir)<}>(p-Tr} = x ...................... (3), 

one of the cutting ellipsoids. 

</> is the same for all because the ellipsoids are similar. 

The plane of intersection of (1) and (3) is found by subtracting 
the equations ; and is therefore 



X. 
The plane of intersection of (2) and (3) is 

+ C X. 



The former of these planes is perpendicular to <j>ir and the latter 
to <f>tr <}ia ; and, since by the question, the former is perpen- 
dicular to the latter, <f>ir is perpendicular to </>TT </>ct, 

.'. S(j>ir (</>TT <a) = 0, 
the equation of the locus of the centres of the cutting ellipsoids. 

This equation will be reduced to the requisite form by ob- 
servin that 



.'. S (IT - a) < 2 7T = 0, 

the equation of an ellipsoid of which the semi-axes are propor- 
tional to 

a 2 , b\ c* (63. 6). 

The Cartesian equation is 



142 QUATERNIONS. [CHAP. VIII. 

Ex. 14. If a tangent plane be drawn to the inner of two 
similar concentric and similarly situated ellipsoids the point of 
contact is the centre of the elliptic section of the outer ellipsoid. 

Let Sp<f>p - 1 be the equation of the inner, 

a*Sp<f>p = 1 of the outer ellipsoid. 
The tangent plane is STT^P = 1. 

Now if a- be the vector to the elliptic section measured from 
the point of contact, IT p + cr is a point in the outer ellipsoid ; 

.'. a 2 S (p + cr) < (p + cr) = 1. 
But crc/>p = (57. Cor.); 



the equation of an ellipse of which the centre is the point of 
contact, 

Ex. 15. find the equation of the curve described by a given 
point in a line of given length whose extremities move in fixed 
straight lines. 

First, let the straight lines lie in one plane. 
Let unit vectors parallel to them be a, ft. 

Let the vectors of the extremities of the moving line be 
xa, y/3, and its length I. Then the condition is 



or x* + y 2 + 2xySoip = l 3 (1). 

The vector to a point which divides this line in the ratio 

e : 1 is 

p = xa + e (yfi xa) 

= xa (1 - e) + ey/3 ; 
. '. Sap = - (1 - e) as + eySa(3, 
= (1 - e) xSa/3 - ey ; 



ART. 64.] CENTRAL SURFACES OF THE SECOND ORDER. 143 

, Sap + SaBSBp SBp + SaBSao 
whence X ~T\ \ /tr* o i \ > y = :-, 

which values being substituted in equation (1) give the required 
equation, viz. : 

(Sap + SapS Pp) a 

+ 2 a " (Sap + SapSpp) (SP P + 
e(L-e) 

= P (S 2 ap - 1)*. 

But p is subject to the additional condition (31. 2. Cor. 2) 
S . app = ; and the locus is a plane ellipse. 

When the given straight lines are at right angles to one 
another, the equation is much simplified, for 

O 1 O A . 

and our equations are 

x 2 + y 2 = I 2 , 

Sap = -(l-e)x, Sp P = -ey; 
whence 



an ellipse of which the semi-axes are le and I (1 e). 

Generally, if the given lines do not meet, let the origin be 
chosen midway along the line perpendicular to both; then we 
have 



y and y being the vectors perpendicular to the lines, 

P = (y+xa)(l-e) + e(-y + yp). 
The first gives 

^ + ( Xa -y^ = -P- 

and the second gives, as in the simpler case above, 
Sap = (le)x+ eySafi, 
= (l-e) xSap - ey. 



144 QUATERNIONS. [CHAP. VIII. 

Hence the elimination of x and y again leads to the equation 
of an ellipsoid, the only difference being that I 2 is diminished by 
the square of the shortest distance between the lines; i.e. the 
axes are less than in the former case. 

In the extreme case, where I = 2Ty, the equation cannot be 
satisfied except by 

x = 0, y = 0, 
(i. e. the locus is reduced to a single point), unless indeed we have 

o = *& 
for then x = y, 

and the locus is a straight line parallel to each of the preceding 
lines. 

65. The cone. 

1. To find the equation of a cone of revolution whose vertex 
is the origin 0. 

Let a be a unit vector along the axis OA, 

p the vector to a point P on the surface of the cone ; 
then Sap = Tp cos 0, 

being the angle POA. 

But this angle is constant, 

.*. S 2 ap - c 2 p 2 is the equation required. 

2. The equation of a cone which has circular sections, but 
which is not necessarily a cone of revolution, is thus found. 

Take the vertex as the origin, and let one of the circular 
sections be the intersection of the plane 

Sap = -a' (1) 

with the sphere p' = Sfip (2). 

Since these are scalar equations we may multiply them together ; 
and thus obtain at all the points of the circular section 

0..... (3). 



ART. 66.] CENTRAL SURFACES OF THE SECOND ORDER. 145 

Now if xp or p be written in place of p, the equation is not 
changed, since p occurs twice on each side. It is therefore the 
required equation, of the cone. 

COR. 1. Every section by a plane parallel to Sap = - a 2 is a 
circle. 

For the equation of a plane parallel to 

Sap = a 2 

is Sap aa 2 , 

which being substituted in the equation of the cone gives 



the equation of a circle. 

COR. 2. The plane S(3p = -bp* ........................... (4) 

also gives a circle whose equation is 

a 2 p 2 = b/3 2 Sap .......................... (5). 

These two equations give the subcontrary sections. 

To deduce the relation between the two sections ; let be the 
vertex of the cone, OAB the plane through a, ft; AB the line in 
which the section cuts this plane, AD that in which the sub- 
contrary section cuts it ; 

OA = P , OB = p, OD = xp. 

6/8 2 
We have, by (5), xp' 2 = ~- 8 - Sap' 



= P 2 ,by(2); 
i.e. OB:OD = OA 2 , 

and the triangles OAB, OAD are similar, or AD cuts OA at the 
same angle that AB cuts OB. 

66. If <fr> - 2a 2 p + aSfip + fiSap, 

the equation of the cone is reduced to 

Sp<j>p = 0. 
T. Q. 10 



146 QUATERNIONS. , [CHAP. VIII. 

It is evident that all the properties of <f>p, Ai-t. 44, are appli- 
cable here. 

As in Art. 57, the equation of the tangent plane is 

0. 



67. EXAMPLES. 

Ex. 1. Tangent planes are drawn to an ellipsoid from a given 
external point, to find the cone which has its vertex at the origin 
[the centre of the ellipsoid], and which passes through all the points of 
contact of the tangent planes with tlie ellipsoid. 

Let a be the vector to the external point, p a point in the 
ellipsoid where a tangent plane through a touches it. 

Then the equation of the ellipsoid is 



and the equation of the tangent plane 

The equation 

Sp<f>p = 

O 2 2 y / V /vO 

/y* ni* iy / Vy/yi If)! && \ * 

mf V if i <* ' ' if (/ *w<w \ 

rt** i " i I ^-L*^* 7 J 

Wl 212 2 L2 ^J 

a o c \a o c / 

represents a surface passing through the points of contact; and 
is the cone required. [For it is homogeneous in Tp."] 

Ex. 2. Of a system of three rectangular vectors two are con- 
fined to given planes, to find the surface traced out by the third. 

Let TT, p, a- be the three vectors, of which two are confined to 
given planes whose equations are 



to find the locus of <r. 

Since the vectors are at right angles, we have 

Sirp = 0, Sir(r 0, Sap = 0, 
and we have five equations from which to eliminate TT and p. 

Since SO.TT = 0, SO-TT = 0, 

IT is at right angles to both a and <r, and therefore to the plane 
off' } or 

IT x Fa<r. 



ART. 67.] CENTRAL SURFACES OF THE SECOND ORDER. 147 

Since Sfip = 0, Sa-p = 0, 

p is at right angles to the plane /?cr; therefore 



and irp = xy Va<r V(3(r. 

Now S-rrp = 0, 

therefore S . Vaa- F/?<r = 0, 

or S (aa- - Sao-) (p<r - Spa) = 0, 

or o*Sap- ScurSp<r=0, 

the equation of a cone of the second order, which has circular 
sections (65. 2). 

COR. The circular sections are parallel to the two planes to 
which the two vectors are confined. 

Ex. 3. The equation p = t s a + u*@ + (t + uf y is that of a cone 
of the second order touched by each of the three planes through 
OAB, OBC, OCA; and the section ABC through the extremities of 
a, (3, y is an ellipse touched at their middle points by AB, BC, CA. 

1. If the surface be referred to oblique co-ordinates parallel 
to a, j8, y respectively, we shall have 

p = xa + yfi.+ zy, 

therefore x = t s , y = u 2 , z = (t + u) 3 , 

or z = ( t jx + l jy) 2 = x + y + 2*jxy, 

which gives (z-x y}* = 4xy, 

a cone of the second order. 

2. If t - u, the equation becomes 

p = t*(a + P), 

the equation of a straight line bisecting the base A, which since 
it satisfies the equation relative to t, shews that this line coincides 
with the cone in all its length; i.e. the cone is touched in this 
line by the plane OAB. 

Similarly, by putting t 0, u - respectively, we can shew 
that the cone is touched by the plane BOC, COA in the lines 
which bisect AC, CA. 

102 



QUATERNIONS. [CHAP. VIII. 

3. Restricting ourselves to the plane ABC, we have the 
section of a cone of the second order enclosed by the triangle 
ABC, which triangle is itself the section of three planes each of 
which touches the cone. 



Ex. 4. The equation p = aa + b/3 + c-y with the condition 
ab + be + ca = is a cone of the second order, and the lines OA, OB, 
00 coincide throughout their length with the surface. 

1. It is evident that the equation gives 

xy + yz + zx = 0. 

2. That if b = 0, c - 0, the question is satisfied by 

p = aa, 
whatever be a, therefore &c. 

Ex. 5. Find the locus of a point, the sum of the squares of 
whose distances from a number of given planes is constant. 

Let AS'8 ] p 1 = C' 1 , S8 2 p 2 = C 2 , &c. be the equations of the given 
planes, p the vector to the point under consideration; then 01,8,, 
x, 8 , &c. will be the perpendiculars on the planes from the point ; 
provided 



therefore SS 1 (p + oj.S,) = C l , &c. 

and ajjSj 2 = C l SS^, &c., 

<V = (C' 1 ->$V) 2 ; 

i.e. the square of the line perpendicular to the first plane from 
the given point 

/C.-^pV 
" V Z'8, ) ' 

and, by the question, 

C.-S8 lP \' /C-SS 2 p\ 2 
' - + ~ - +&c. is constant. 



The locus is therefore a surface of the second order. 



ART. 68.] CENTRAL SURFACES OF THE SECOND ORDER. 149 

Ex. 6. The lines which divide proportionally the pairs of 
opposite sides of a gauche quadrilateral, are the generating lines 
of a hyperbolic paraboloid. 

Let ABCD be the quadrilateral. 
AD, EG are divided proportionally 
in P and R. 

Let CA = a CB = P, CD = ; 



i. e. CP y = m (a y) ; 

therefore RP = CP CR = 




= m(3 +p {y + m (a - y) ra/3} 
therefore x = pm, y = m pm, z = p (1 m); 



therefore 



m=x+ y> ^ = ^v 



x + y 

or (x + z) (x + y) = x, 

the equation referred to oblique co-ordinates parallel to a, /?, y. 

PASCAL'S HEXAGRAM. 

68. Let be the origin, OA, OB, OC, OD, OE five given 
vectors lying on the surface of a cone, and terminated in a plane 
section of the cone ABCDEF, not passing through ; OX any 
vector lying on the same surface. 

Let OA = a, OB = fi, OC = y, OD=8, OJ = e, OX '= p. 

The equation 

S. F(FaF8e) F(F^yFep) F(FySFpa) = (1) 

is the equation of a cone of the second order whose vertex is 
and vector p along the surface. For 



150 QUATERNIOXS. [CHAP. VIII. 

1. It is a cone whose vertex is because it is not altered 
by writing xp for p. Also it is of the second order in p, since p 
occurs in it twice and twice only. 




2. All the vectors OA, OS, OC, OD, OE lie on its surface. 

This we shall prove by shewing that if p coincide with any 
one of them the equation (1) is satisfied. 

If p coincide with a, the last term of the left-hand side of the 
equation, viz. Vpa, becomes Vaa = Va 3 = 0, and the equation is 
satisfied. 

If p coincide with ft, the left-hand side of the equation be- 
comes 

S. F(FaFSe) F(F/? r Fe) F(FySF/3a) (2). 

Now F(Fy3yFej3) = - F(Fe/3F/3y), (22. 2), is a vector parallel 
to /? (31. 3), call it mp; and 
F.{F(Fa/3FSe) F(FySF/2a)} = F. {F(Fa/3FSc) F(Fa/?FyS)}, (22. 2), 

= a multiple of Ya{3, (31. 3), 

= nVa{3, say. 



ART. 68.] CENTRAL SURFACES OF THE SECOND ORDER. 151 

Hence the product of the first and third vectors in expression 
(2) becomes 

scalar + n Fa/3, 

and the second is m/2; therefore expression (2) becomes, by 31. 2, 
$ . (scalar + n Fa/3) m/2 



= 0, 
because Fa/2 is a vector perpendicular to (3. 

Equation (1) is therefore satisfied when p coincides with ft. 

If p coincide with y both the second and third vectors are 
parallel to /3 (31. 3); therefore their product is a scalar, and equa- 
tion (1) is satisfied. 

The other cases are but repetitions of these. 

Hence equation (1) is satisfied if p coincide with any one of 
the five vectors a, /3, y, 8, e; i.e. OA, OB, OC, OD, OE are vectors 
on the surface of the cone. 

3. Let F be the point in which OX cuts the plane ABCDE; 
then ABCDEF are the angular points of a hexagon inscribed in 
a conic section. 

4. Let the planes OAB, ODE intersect in OP; OBC, OEF 
in OQ; OCD, OF A in OR', then 

V. Va(BV8e = mOP, (31. 4), 
F. 

V. 

therefore 

S.V(Va/3 FSe) F ( F/2y Fcp) F ( FyS Fpa) = mnpS(OP .OQ.OR}; 
hence equation (1) gives 

8(OP.OQ.OR)=Q, 
or (31. 2. Cor. 2) OP, OQ, OR are in the same plane. 

Hence PQR, the intersection of this plane with the plane 
A BGDEF is a straight line. But P is the point of intersection 
of AB, ED, &c. 



152 QUATERNIONS. [CHAP. VIII. 

Therefore, the opposite sides (1st and 4th, 2nd and 5th, 3rd 
and 6th) of a hexagon inscribed in a conic section being produced 
meet in the same straight line. 

COR. It is evident that the demonstration applies to any six 
points in the conic, whether the lines which join them form a 
hexagon or not. 

ADDITIONAL EXAMPLES TO CHAP. VIII. 

1. Find the locus of a point, the ratio of whose distances 
from two given straight lines is constant. 

2. Find the locus of a point the square of whose distance 
from a given line is proportional to its distance from a given 
plane. 

3. Prove that the locus of the foot of the perpendicular from 
the centre on the tangent plane of an ellipsoid is 

(axy + (byy + (czy=(x* + y* + z 2 y. 

4. The sum of the squares of the reciprocals of any three 
radii at right angles to one another is constant. 

5. If Qy v Oy a , Oy a be perpendiculars from the centre on 
tangent planes at the extremities of conjugate diameters, and if 
Qu Q& Q 3 be the points where they meet the ellipsoid; then 

1 1 11)1 

1 . -f . = 1 1 . 

OY 2 00 2 OY 2 00 OY a 00 ct* b* c 

6. If tangent planes to an ellipsoid be drawn from points in 
a plane parallel to that of xy, the curves which contain all the 
points of contact will lie in planes which all cut the axis of z 
in the same point. 

7. Two similar and similarly situated ellipsoids intersect 
in a plane curve whose plane is conjugate to the line which joins 
the centres of the ellipsoids. 

8. If points be taken in conjugate semi-diameters produced, 
at distances from the centre equal to p times those semi-diameters 
respectively; the sum of the squares of the reciprocals of the 



ART. GS.] CENTRAL SURFACES OF THE SECOND ORDER. 153 

perpendiculars from the centre on their polar planes is equal to p z 
times the sum of the squares of the perpendiculars from the 
centre on tangent planes at the extremities of those diameters. 

9. If P be a point on the surface of an ellipsoid, PA, PB, 
PC any three chords at right angles to each other, the plane 
ABC will pass through a fixed point, which is in the normal to 
the ellipsoid at P; and distant from P by 

2 

P 

!_ 1_ i' 
a* b 2 c 2 

where p is the perpendicular from the centre on the tangent 
plane at P. 

10. Find the equation of the cone which has its vertex in 
a given point, and which touches and envelopes a given ellipsoid. 



CHAPTER IX. 

FORMULAE AND THEIR APPLICATION. 

69- PRODUCTS of two or more vectors. 

1. Two vectors. The relations which exist between the 
scalars and vectors of the product of two vectors have already 
been exhibited in Art. 22. We simply extract them : 

(a) Sap = Sp a . (b) Va/3 = -Vj3a. 

(c) ap + pa=2Sap. (d) a/? - /3a = 2 Fa/?. 

These we shall quote as formulae (1). 

2. We may here add a single conclusion for quaternion 
products. 

Any quaternion, such as aft, may be written as the sum of 
a scalar and a vector. If therefore q and r be quaternions, we 
may write 



r = Sr+Vr; 

qr = SqSr + SqVr + Sr Vq + Vq Vr, 
S . qr = SqSr +S.Vq Vr, 
V. qr = SqVr + SrVq + V. VqVr, 

where S .VqVr is the scalar part, and V.VqVr the vector part of 
the product of the two vectors Vq, Vr. 

If now we transpose q and r, and apply (a) and (b) of for- 

mulae 1, we get 

S.qr = S.rq \ 

V. qr + V. rq=2 (SqVr + SrVq)) " 



ART. G9.J FORMULA AND THEIR APPLICATION. 155 

3. Three vectors. By observing that S.ySafi is simply the 
scalar of a vector, and. is consequently zero, we may insert or 
omit such an expression at pleasure. By bearing this in mind 
the reader will readily apprehend the demonstrations which 
follow, even in cases where we have studied brevity. 



-S.yap .............................. (3). 

Again, S.a{3y = S.a (S/3y + Vpy) 



(3). 

The formulae marked (3) shew that a change of order amongst 
three vectors produces no change in the scalar of their product, 
provided the cyclical order remain unchanged. 

This conclusion might have been obtained by a different pro- 
cess, thus : 

In (2) let q - a/2, r = y, there results at once 



Again in (2) let q = ya, r = ft, there results 

S . yap = S . Pya. 
We have therefore, as before, 

S.apy=S.yap = S.pya .................... (3). 

4. S.afiy^S .aVfiy 

= -S.aVy(3, (by 1.6), 

= -S.arf ............................. (4). 

Similarly S . a(3y = - S . fay ........................ (4), 

or a cyclical change of order amongst three vectors changes the 
sign of the scalar of their product. 



156 QUATERNIONS. [CHAP. IX. 

5. It has already been seen (Art. 31. 1) that S . afiy is the 
volume of the parallelepiped of which the three edges which 
terminate in the point are the lines OA, OS, OC whose vectors 
are a, (3, y respectively. 

We may express this volume in the form of a determinant, 
thus : 

Let a, /?, y be replaced by 

xi +yj + zk, x'i + y'j + z'k, x"i + y"j + z"k (Art. 31. 5) ; 
x, y, z being the rectangular co-ordinates of A, x, y, z' those of B, 
x", y", z" those of C, measured from as the origin ; then 

S . a(3y = S . (xi + yj + zK) 
x (x'i + y'j + z'k) 
x (x"i + y"j + z"k). 

Now if we observe first that the scalar part of this product is 
confined to those terms in which all the three vectors i, j, k 
appear ; and secondly that the sign of any term in the product 
will by formulae (3) and (4) be or + according as cyclical order 
is or is not retained, we perceive that we have the exact con- 
ditions which apply to a determinant : therefore 

S . a{3y = - | , y , z 

x, y , z' 



.(5). 

r* n f* 

x , y , z 
The volume of the pyramid OABC is one-sixth of the above. 

Note relative to the sign of the scalar. 

Since ijk = - 1 (19), it is clear that if OA, OB, OC assume the 
positions of Ox, Oy, Oz in the figure of Art. 16, S (OA . OB . OC) 
will have a minus sign, whilst the order of the letters A, B, C is 
right-handed as seen from 0. 

If now we take any pyramid whatever OABC, of which the 
vertex is 0, and assume that S (OA . OB . OC) (which, being pro- 
portional to the volume of the pyramid, we may designate OABC), 
is negative when the order of the letters A, B, C is right-handed 



ART. C9.] FORMULAE AND THEIR APPLICATION. 157 

as seen from 0, we shall find the following general law of signs to 
hold good whatever be the vertex ; viz. the sign of the scalar is 
minus or plus according as the order in it of the angles of the base 
of the pyramid is right-handed or left-handed as seen from the 
vertex. 

For example, CA BO = S (CA . CB . CO] 



- - Sapy 

= - OAC, 

which is plus because OABC is minus, and the order of the letters 
A, B, as seen from C is left-handed. 

6. V.a = V. 



=aSpy-V.aVyp,(l.b), 
.b), 

(6). 



7. V.apy = r.(Sap+ 

= ySafi -V. yVa/3 ; 



therefore F. ay+F. ya;8=2ySaj8 ....................... (7). 

8. 2r.a 



= F. apy + V. ya{3 - ( F. ay/3 + F. ya,3) 
-F(aj8y + ^ay)-F(ay/? + ya^), (by 6), 
=F. (op + Pa)y- V. (ay + ya) p 
= 2ySap-2pSay, (1. c); 

therefore F. aFySy = 7 Sap-pSay ....................... (8). 



158 QUATERNIONS. [CHAP. IX. 

9. We have, by (8), 



therefore, by addition, 

V.(aVpy+pVya + yYap) = Q .............. (9). 

10. F. apy = F. a (Spy + Vpy) 



which, by (8), = aSpy - pSay + ySap ........... (10). 

Another proof of this important formula is found in the 
identity 



which, by (4) and (6), is the theorem itself. 

11. If in (8) we write Fa/3 in place of a, we get 
V. VaV 



= ~PS.apy ........................... (11). 

12. Four vectors. If in (8) we write FaS in place of a, we 
obtain 

V(Va8Vpy) = yS.a8p-pS.a8y ............ (12). 

13. By (12) we have 

F ( Vpy FaS) - 8S . Pya - aS . py8. 
But F ( Vpy FaS) - - F ( FaS Vpy). 

Hence, by adding the above result to (12), we get 
SS. pya - aS . PyS + yS . a8p - pS . aSy = 0, 

which, by (3) and (4), if we adopt alphabetical order, may be 
written 

aS.py8-pS.a.yS + yS.apS-8S.apy=Q ...... (13), 

or 8S.apy = aS.py8-pS.ay8 + yS.ap8 .......... (13), 



ART. CO.] FORMULA AND THEIR APPLICATION. 159 

or, again, if we adopt cyclical order, 

aS . fty8 - 8S . afty + yS. Baft - ftS . ySa, 
or, finally, SS. afty = aS . ftyS-ftS.y8a + yS. Saft ........ (13). 

This equation expresses a vector in terms of three other 
vectors. The following equation expresses it in terms of the 
vectors which result from their products two and two. 

14. F(ySa/3) may be written, first as F(y . Sa/3), and secondly 
as F(yS. aft), and the results compared. These forms give re- 
spectively 

V (y . Sa/3) = F. y (S. Sa/3 + F . Sa/3) 

= yS . a/3S + F. y (SSa/3 - aSSft + ftSSa), by (3) and (10), 

- yS . a/2S + VySSaft - VyaSSft + VyftSBa ; 
F (yS . aft) = V . (SyS + FyS) (Sap + Fa/3) 

- VaftSyS + VySSaft+ F. FySFa 

- VapSy&+ VyZSaft- F. Fa/3 FyS 

= Fa/3SyS + FySSa/3 - BS . afty + yS . a/3S, by (12). 

The two expressions being equated, and the common terms 
deleted, there results 

SS.afty= VaftSy8+ VftySaS + VyaSpS ......... (14). 

15. S.afty8 = S.(S.afty+V.afty)S 

= S.(V.afty)5 

= S . (aSfty - ftSay + ySaft) 8, by (10), 

= SaftSyS-SaySft$ + Sa$Sfty ............... (15). 

16. S ( Vaft FyS) = S . (op - Saft) (yS - .S'yS) 



= SaSSpy - SaySftS, by (15) ......... (16). 

17. S.aftyS=S.(Vafty)8 

= S.8Vafty 
= S.Sap y ......................... (17). 



1GO QUATERNIONS. [CHAP. IX. 

18. Five vectors. As we do not purpose to exhibit any 
applications of the relations which exist among five or more 
vectors, we shall confine ourselves to simply writing down the two 
following expressions. 



F.a/?ySe = V. Sya .................... (18). 

70. Many of these formulae might have been proved differ- 
ently, and some of them more directly, by assuming for instance 
that a, (3, y are not in the same plane. In this case any other 
vector 8 may be expressed in terms of a, (3, y, by the equation 

S = xa + yp + zy, (31. 5); 
therefore S./3yS = xS . pya = xS . a/3y, (3) , 



S.$a/3 = zS . yap = zS . a/3y, (3) ; 
therefore ?>S . a/3y = xaS . ajBy + yftS . apy + zyS . afty 

-- aS . PyS - pS . ySa + yS . Sa/3 
which is formula 13. 

71. EXAMPLES. 

Ex. 1. To express the relation between the sides of a spherical 
triangle and the angles opposite to them. 

Retaining the notation and figure of Ex. 2, Art. 29, we shall 
have 

Fa/3 Vpy = y' sin c . a sin a, 

where y, a are unit vectors perpendicular respectively to the 
planes OAB, OBC. 

Therefore F . Fa/2 F/?y = sin c sin a . /? sin E. 

Also -pS.apy = P sin c sin </>, (31. 1), 

where < is the angle between OC and the plane OAB. 

Now these results are equal (formula 11), therefore 
sin < = sin a sin . 



ART. 71.] FORMULAE AND THEIR APPLICATION. 161 

Similarly sin (ft sin b sin A ; 

therefore sin a sin E = sin b sin A, 

or sin a : sin b :: sin A : sin B. 

Ex. 2. ^o find the condition that the perpendiculars from the 
angles of a tetrahedron on the opposite faces shall intersect one 
another. 

Let OA, OB, 00 be the edges of the tetrahedron (Fig. of Art. 
31), a, /?, y the corresponding vectors. 

Vector perpendiculars from A and B on the opposite faces are 
F/3y, Fya respectively (22. 8). If these perpendiculars intersect 
in G, the three points A, B, G will be in one plane, whence 

S.(ft-a) F/3yFya = (31. 2, Cor. 2), 
i.e. S.(/3-a)V. FySyFya-0. 

Now F . Fy Fya = - yS . jSya (Formula 1 1), 

therefore S . (ft - a) F . Fy3y Fya = - (S/?y - Say ) S . jffya. 

Hence Spy = Say. 

Now 5(7' + a4' = -S a + a 2 



= a 4- + y - 

= a 2 + /3 2 + y 2 - 2Say 

= (7-) 2 + yS' 
- AC* + OB*. 

Consequently the condition that all three perpendiculars shall 
meet in a point is that the sum of the squares of each pair of 
opposite edges shall be the same. 

COR. Conversely, if the sum of the squares of each pair of 
opposite edges is the same, the perpendiculars from the angles on 
the opposite faces will meet in a point. 

Ex. 3. If P be a point in the face ABC of a tetraliedron, 
from which are drawn Pa, Pb, PC, respectively parallel to OA, 
OB, OC to meet the opposite faces OBC, OCA, OAB in a, b, c; 
then will 

Pa Pb_ jPc 
OA + OJJ + ~OC~ 
T.Q. 11 



162 QUATERNIONS. [CHAP. IX. 

Retaining the notation of the last examples, let OP = 8, 
Pa = tea, Pb = y(3, PC = 2y ; then 

\Jdi O " OCGL \JO == O ~~* 2//5* C/C ~- = - O ^ *Y" 

Now because P, A, B, C are in the same plane 
and because 0, a, B, C are in the same plane 

(2); 



also because O, A, b, C are in the same plane 

S.(S-2,/?)ya = 0, 

i.e. yS. fiya = S. Sya, 

or, by formula 3, yS.a(3y = S. Sya ........................ (3); 

lastly, because 0, A, JB, c are in the same plane 

S.(S-zy)a/3 = 0, 

ie. zS . yaj3 = S . 8a(3, 

or zS.a{3y = S. Sa(3 ........................ (4). 

Adding (2), (3), and (4) there results 

. a(3y = S. 8fiy + S. Sya+S. Sufi 



therefore x + y + z = I : 

Pa P& Pc__ 
O4 + OB + OC~ 

COB. 1. If P be in the plane ABC produced below the plane 
OBC, Pet as a vector will have the same sign as OA has; hence 
in this case we shall have 

_Pa L Pb Pc__ 
OA + OB + OC~ 



ART. 71.] FORMULA AND THEIR APPLICATION. 163 

COR. 2. If P be outside both the planes OBC, OCA ; we 
shall have 

Pa,_Pb Pc__ 
~ OA OB + 00 ~ 

Ex. 4. Any point Q is joined to the angular points A, B,C,0 
of a tetrahedron, and ilie joining lines, produced if necessary, 
meet the opposite faces in a, b, c, o ; to prove that 

Qa Qb Qc Qo_ 
Aa Bb Cc Oo 

regard being had to the signs of Aa, Bb, &c., as in the last example. 

Let #4=0, QB = P, QC = y, QO = S; Qa = aa, Qb = bj3, Qc = cy, 
Qo = d8: then since the points a, b, c, o are in the planes 00, 
AGO, ABO, ABC, respectively, we have, as in the last example, 

aS . a (py + y8+Sp) = S. pyo, 

&c. &c. 

i.e. aS.(apy+ay8+aSp)-S.pyS = Q ............. (1), 

bS. (fay + (3y8 + /?8a) - S . ay8=0 ............. (2), 

............. (3), 

Q ............ (4). 

Now, if we write 

S.apy=X, S.ay8 = y, S.aop^z, S.py8 = u; 
and apply the formulae 3 and 4, we get 

ax + ay + az u = 0, 

bx y bz + bu = 0, 

cx + cy+ z cu Q, 



a d 

which give - 1- x + -y ^ u0, 

ct 1 d L 



a 



112 



164 QUATERNIONS. [CHAP. IX. 

c b 

c- 1 y b-\ Z 

c d 



c-1 d-l 

and. therefore, 7 + -= r + , + - 7 , = 0, 

a- 1 6-1 c-1 tt 1 

abed 
i.e. 



:r 7 =- -- , 

a-1 o-l c 1 a-l 

Qa Qb Qc Qo 

or ~ + "zJI + 7T + 7T = 1- 

A a Bb Cc (Jo 

Ex. 5. 7/**H>0 tetrahedra ABCD, A'B'C'D' are so situated that 
the straight lines, A A', BB', CC', DD' all meet in a point, the lines 
of intersection of the planes of corresponding faces shall all lie in 
the same plane. 

Let A' A, B'B, C'C, D'D meet in 0. 



The equation of the plane ABC is (34. 5) 

Sp ( FayS + F/3y + Fya) = S . afiy, 
and that of A'B'C' becomes, after dividing both sides by mnp, 

Sp (- VaB + - Fy + - Fya^ = S . a/?y. 

r \p m n ' / 

The vector line of intersection of the two planes is (34. 9) 
F. (Fa/3 + F/3y + F ya ) Q Fa^-f 1 F/3y + 1 Fya) , 

i.e. by formula (11), omitting the common factor S . afiy, 

/I 1\ /I 1\_ /I IN 
( ---- a + ( ---- /3 + ----- y. 
\n pj \p mj \m nj 

From this expression the vectors of the intersections of the 
other planes may at once be written down. 



ART. 71.] FORMULAE AND THEIR APPLICATION. 165 

That of ABD, A'B'D' is 

/I 1\ /I 1\ /I 1\_ 
( --- a+ --- )/3+( --- )S; 
\n qj \q mj \m nj 

that of AC D, A'C'D'is 

/! l \ f l l \ /I 1\* 
--- )a+( --- y + ( --- )8; 

\p qJ \q mj \m p/ 
and that of BCD, B'C'D' 

/I 1\. /I 1\ /I 1\. 
--- )P + ( --- y+ --- S. 
\p qJ \q / ' \n p/ 

Now to prove that any three of these lines lie in the same 
plane, all that is necessaiy is to prove (31. 2, Cor. 2) that the 
scalar of the product of their vectors equals 0. 

If we take the vectors of the first three, we may write them 
under the form 



b(3 + cy, da. + b'fi + cS, a"a + b'y - bS, 
respectively ; so that the scalar of their product is 

S.(aa + bfi + cy) (a a + b'fi + cS) (a" a. + b'y - 68). 

Now the coefficient of every different scalar in this product is 
separately equal to 0. That of S . a/3y for instance is, omitting 
the common factor b', 



m \m n \p q \p m \n 
in which every term vanishes. 
That again of S . /3yB is 

bcb' + cb'b, 
which is ; and so of the rest. 

Hence the intersections, two and two, of the first three pairs 
of planes lie in the same plane ; and the same may be proved in 
like manner of any other three : whence the truth of the pro- 
position. 



166 QUATERNIONS. [CHAP. IX. 

Ex. 6. CP, CD are conjugate semi-diameters of an ellipse, 
as also CP', CD' ; PP 1 , DD' are joined ; to prove that the area of 
tlie triangle PGP equals that of the triangle DCD'. 

Let a, ft, a', ft' be the vectors CP, CD, CP f , CD' ; k a unit 
vector perpendicular to the plane of the ellipse. 

Since 

a.*=if/~ l \l/a = (aiSi\l/a + bjSj{j/a), &c., &c. (47. 5), 

therefore Vaa= V. (aiStya + bjSjij/a) (aiSitya! -f bjSjij/a!) 
= able (Siif/aSfya SfyajSiij/a) 
= - abkS . k V (i/^a'). (Formula 1 6.) 

Similarly V/3p'=-abkS. kV(^p'). 

Now \l/a, \l/fi are unit vectors at right angles to one another; 
as are also \j/a, tyf? ; therefore the angle between iffa and tya! is 
the same as that between i}//3 and \frfi'. 

Hence S . k V (^a!) = S.kV ($&$&), 

and Vaa.'=Vpp, 

i.e. area of triangle PGP' that of triangle DCD'. 

Ex. 7. If a parallelepiped be constructed on the semi-con- 
jugate diameters of an ellipsoid, the sum of the squares of the areas 
of the faces of the parallelepiped is equal to the sum of the squares 
of the faces of the rectangular parallelepiped constructed on the 
semi-axes. 



By 63. 9, a = - (aiStya + bjSfya + c 
fi = - (cnStyfi + bjSfyP + c 
therefore Fa/3 = abk (StyaSjfyp - S 
+ acj 
+ bci 

Now Si\j/aSj^p-Si^pSj(j/a = SVij7^a, Formula (16), 

, (Art. 17); 



ART. 71.] FORMULAE AND THEIR APPLICATION. 167 

therefore Va.fi = (abkSk^y + acjSfyy + bciSiij/y), 
Vya = - (abkStyp + acjSj^/3 + bciStyP), 
V(3y = (abkSktya. + acjSjif/a + bciSiij/a). 

If now we square and add these expressions, observing that 
because \f/a, if/{3, ij/y are unit vectors at right angles to one another, 

(Stya) 1 + (Styp)' + (Styy)* = 1, 
we shall have 

( Fa/J)* + ( Fay)* + ( Y{3y)> = - {(ab)* + (acf + (6c) 2 }, 
which (21. 4) is the proposition to be proved. 

Ex. 8. To find the locus of t/te intersections of tangent planes 
at the extremities of conjugate diameters of an ellipsoid. 

Let TT be the vector to the point of intersection of tangent 
planes at the extremities of a, ft, y : then 

S7r</>a= 1, (57), 

gives Sirif/'a. = 1, 

or S\l/Tr\l/a.~ 1, 

Silnnjrp = - 1, 

Sif/mj/y = 1. 

From these three equations we extricate i}/ir by means of for- 
mula (14), which gives 



iff IT = ViJ/aif/P + Fi/f/fyy + V\fryt(/a 



= -3, 

a?_ y 2 *_ 

3c? + 36 2 + 3? ~ ; 

an ellipsoid similar to the given ellipsoid. 



1C8 . QUATERNIONS. [CHAP. IX. 

Ex. 9. I/O, A, B, C, D, E are any six points in space, OX 
any given direction, OA', OB', OC', OD', OE' the projections 
o/OA, OB, OC, OD, OE on OX; BCDE, CDEA, DEAB, EABC, 
A BCD the volumes of the pyramids whose vertices are B, C,D,E,A, 
with a positive or negative sign in accordance with the law given 
in the note to 69. 5 ; then 

OA'. BCDE + OB'. CDEA + OC'. DEAB + OD'. EABC 



Let OA, OB, OC, OD, OE be a, /?, y, S, e respectively. 
Write for aS (y -ft) (8- ft) (e - /?) its value 

a ( . ySe - S . S e/ 8 + S . c/?y - . yS), 

and similar expressions for /3 (a y) (S - y) (e y), die., and there 
will result, by addition, 



+ cSG8-)(y-a)(8-a)=O f 

i.e. retaining the notation adopted in the Note referred to, 
OA . BCDE+ OB . CDEA + OC . DEAB + OD . EABC 



Now let ir be a vector along OX ; then the operation by S . TT 
on the above expression gives the result required. 



In some of the examples which follow, we will endeavour to 
shew how a problem should not, as well as how it should, be 
attacked. 

Ex. 10. Given any three planes, and the direction of the vector 
perpendicular to a fourth, to find its length so that they may meet 
in one point. 

Let /Sap = a, Sj3p = b, Syp = c be the three, and let S be the 
vector perpendicular to the new plane. Then, if its equation be 

tSSp = d, 



ART. 71.] FORMULAE AND THEIR APPLICATION. 169 

we must find the value of d that these four equations may all be 
satisfied by one value of p. 

Formula (14) gives 

pS . apy = VafiSyp + VfiySap + VyaSfip 



by the 'equations of the first three. Operate by S . 8, and use the 
fourth equation, and we have the required value 

dS . afiy = aS . fiyo + bS . yaS + cS . afiS. 

Ex. 11. The sum of the (vector) areas of the faces of any 
tetrahedron, and therefore of any polyhedron, is zero. 

Take one corner as origin, and let a, ft, y be the vectors of 
the other three. Then the vector areas of the three faces meeting 
in the origin are 

- Fa/?, - F/3y. - Fya, respectively. 

a a m 

That of the fourth may be expressed in any of the forms 

lF( 7 -a)(/3-a), lF(a-/?)(y-/?), ' 

But all of these have the common value 



which is obviously the sum of the three other vector- areas taken 
negatively. Hence the proposition, which is an elementary one in 
Hydrostatics. 

Now any polyhedron may be cut up by planes into tetrahedra, 
and the faces exposed by such treatment have vector-areas equal 
and opposite in sign. Hence the extension. 

Ex. 12. If the pressure lie uniform throughout a fluid mass, 
an immersed tetrahedron (and therefore any polyhedron) experiences 
no couple tending to make it rotate. 

This is supplementary to the last example. The pressures on 
the faces are fully expressed by the vector-areas above given, and 



170 QUATERNIONS. [CHAP. IX. 

their points of application are the centres of inertia of the areas 
of the faces. The co-ordinates of these points are 

<+0. J(/* + y)> |(y + ), l( + P + y), 

and the sum of the couples is 

I V. [Yap. (a + /3) + F/3y.(/3 + y) + Fya. (y+a) 
+ F(y/3 + fia + ay).(a + /3 + y)} 
= -| F(Fa/?. y + F/?y . a+ Fya . /3) = 0, 

by applying formula (9). 

Ex. 1 3. What are the conditions that the three planes 

Sap = a, S(3p = b, Syp = c, 
shall intersect in a straight line ? 

There are many ways of attacking such a question, so we will 
give a few for practice. 

(a) pS . afiy = VafiSyp + VfiySap + YyaS/Bp 

= cVafi + aVfiy + bVya 

by the given equations. But this gives a single definite value 
of p unless both sides vanish, so that the conditions are 

.a/3y=0, 

and c Fa/? + a Vfiy + b Fya - 0, 

which includes the preceding. 

(b) S (la - m/3) p = al-bm 

is the equation of any plane passing through the intersection of 
the first two given planes. Hence, if the three intersect in a 
straight line there must be values of I, m such that 

la m(3 y, 
la rnh = c. 

The first of these gives, as before, 



ART. 71.] FORMULA AND THEIR APPLICATION. 

and it also gives 

Vya = m Fa/3, Vpy = -l Fa ft, 

so that if we multiply the second by Fa/?, 

la Vap - mb Fa/3 = c Fa/3 

becomes a F/3y b Vya = c Fa/3 ; 

the second condition of (a). 

(c) Again, suppose p to be given by the first two in the form 

p = pa + qP + X Fa/3, 

we find a = pa* + qSap, because /So. Fa/3 = 0, 

6 =pSap + qp 2 ; 



therefore 



a 8 , Sap 

Sap, p* 



a, Sap 
b, P 2 



a , a 

Sap, b 



so that the third equation gives, operating by S . y, 



a 2 , Sap 
Sap, p 



a, Sap 



a , a 
Sap, b 



. a/3y. 



Now a determinate value of x would mean intersection in one 
point only ; so, as before, 

C (a*P* - S*ap) = a (p 2 Say - SapSpy) - b (SapSay- a'Spy). 

The latter may be written 

S.a[c (a/3 2 - pSap) - a (y/3 2 - pSpy) - b (aSpy - ySap)] - 0. 
S. a(ap 3 -pSap) = Sa(p.pa- i 



= -S.a(p Yap) =-S (a/3 Fa/3). 
Similarly, S . a (y 2 - pSpy) --= S (aft Vpy), 
and S . a (aSpy - ySap) = S.a(V.p Fya), (formula 8), 

= S (a/3 Fya). 
The equation now becomes 

S . ap (c Fa/3 + a F/3y + b Fya) = 0. 



172 QUATERNIONS. [CHAP. IX. 

Now since S . a/3y = 0, a, (3, y are vectors in the same plane; 
therefore y may be written ma + nfi, 

and c Fa/3 4- a F/3y + b Fya 

assumes the form Fa/2, which, unless e = 0, gives 
(a/3 Fa/3) = 0, 

or Fa/3 is in the same plane with a, ft; but it is also perpendicular 
to the plane, which is absurd ; therefore e = 0, or 

cVa/3 + aF/3y + 6 Fya = ; 

thus the third and prolix method leads to the same conclusion as 
the first. 

Ex. 14. Find the surface traced out by a straight line which 
remains always perpendicular to a given line while intersecting 
each of two fixed lines. 

Let the equations of the fixed lines be 

nr = a 4- rr/3, w l = a t 4- xfi^ 

Then if p be the vector of the new line in any position, 
p = iff + y (ra'j OT) 



This is not, as yet, the equation required. Fur it involves 
essentially three independent constants, x, x lt y ; and may there- 
fore in general be made to represent any point whatever of 
infinite space. The reader may easily see this if he reflects that 
two lines which are not parallel must appear, from every point of 
space, to intersect one another. We have still to introduce the 
condition that the new line is perpendicular to a fixed vector, 
y suppose, which gives 

S. 7 (K 1 -vr) = Q = S. 7 [(a l -a) + x i P l -xp]. 

This gives x l in terms of x, so that there are now but two 
indeterminates in the equation for p, which therefore represents 
a surface, which, it is not difficult to see, is one of the second 
order. 



ART. 71.] FORMULAE AND THEIR APPLICATION. 173 

Ex. 15. Find the condition that the equation 

s.pfr^i 

may represent a surface of revolution. 

The expression <frp here stands for something more general than 
that employed in Chap. VIII. above, in fact it may be written 



where a, a p (3, /?,, y, y t are any six vectors whatever. This will 
be more carefully examined in the next chapter. 

If the surface be one of revolution then, since it is central 
and of the second degree, it is obvious that any sphere whose 
centre is at the oi-igin will cut it in two equal circles in planes 
perpendicular to the axis, and that these will be equidistant from 
the origin. Hence, if r be the radius of one of these circles, e the 
vector to its centre, p the vector to any point in its circumference, 
it is evident that we have the following equation, 



where C and e are constants. This, being an identity, gives 



The form of these equations shews that C is an absolute con- 
stant, while r and e are related to one another by the first ; and 
the second gives 

(j>p Cp + e/Sep. 
This shews simply that . ep</>p = 0, 

i. e. c, p, and <p are coplanar, i. e. all the normals pass through a 
given straight line ; or that the expression 

Vp<f>p, 

whatever be p, expresses always a vector parallel to a particular 
plane. 

Ex. 16. If three mutually perpendicular vectors be drawn 
from a point to a plane, the sum of the reciprocals of the squares 
of their lengths is independent of their directions. 



174 QUATERNIONS. [CHAP. IX. 

Let Sep = 1 

be the equation of the plane, and let a, (3, y be any set of 
mutually perpendicular unit-vectors. Then, if xa, yf$, zy be 
points in the plane, we have 

= 1, ySpe = 1, zSyf = 1, 



whence - - oSae + pSQe + ySye (63. 2) = - + + Z . 
Taking the tensor, we have 

.i. + i + i 

** * 2 2 2 

x y z 

Ex. 17. Find the equation of the straight line which meets, 
at right angles, two given straight lines. 

Let CT = a + xft, ro- = Oj 4- a; 1 ^ 1 , 

be the two lines ; then the equation of the required line must be 
of the form 



and nothing is undetermined but a 2 . 

Since the first and third equations denote lines having one 
point in common, we have 



Similarly S . ft FySft (a, - a 2 ) = 0. 

Let *, = yP+yA 

(it is obviously superfluous to add a term in F/?ft), then 

s. 

8.+ 

and, finally, 



Ex. 1 8. IfTp=Ta=Tp = l, and S.a/3 P = 0, 
^.^(p-a)^(p- ) 8) =:/v /i(l-^). 
Interpret this theorem geometrically. 



ART. 71.] FORMULAE AND THEIR APPLICATION. 175 

We have, from the given equations, the following, which are 
equivalent to them, 



p = xa 
Hence -x 2 -7/* + 2xySap = - 1, 

U( \- ( 

* 



S.U(p-a}U(p~P) 



-2(xy-x)Sap 



_x+y-\ / I -Sa/3 

2 V l-x-y + xy(l+Sa(3) 

= x + y-l / I -Sap 

2 V l-x-y + % (2xy + x* + y* - I) 

= x^-y-l. f~ 

2 V l-2 



-Sap 



Of course there are far simpler solutions. Thus, for instance, 
the given equations shew that p, a, p are radii of some unit 
circle. Hence the expression is the cosine of the supplement of 
the angle between two chords of a circle drawn from the same 
point in the circumference. This is obviously half the angle 



176 QUATERNIONS. [dlAP. IX. 

subtended at the centre by radii drawn to the other ends of the 
chords. The cosine of this anle is 



and therefore the cosine of its half is 



v^ j 

Ex. 19. Find the relative position, at any instant, of two 
points, which are moving uniformly in straight lines. 

If a', ft be their vector velocities, t the time elapsed since 
their vectors were a, ft, their relative vector is 

p = a + t of - ft - tft' 

so that relatively to one another the motion is rectilinear, and 
the vector velocity is 

a'-/?. 

To find the time at which the mutual distance is least. 
Here we may write 

Tp* = -y 2 -2tSyo-t>? 



As the last term is positive, this is least when it vanishes, 
Le. when 

t = -S.y8- 1 . 
This gives p = y S/S'yS" 1 

= 7 F8-'y, 

the vector perpendicular drawn to the relative path; as is, of 
course, self-evident. 

Ex. 20. Find the locus of a given point in a line of given 
length, when the extremities of the line move in circles in one plane. 
(Watt's Parallel Motion.) 



ART. 71.] FORMULA AND THEIR APPLICATION. 177 

Let a- and r be the vectors of the ends of the line, drawn 
from the centres a, ft of the circles. Then if p be the vector of 
the required point 



subject to the conditions 



From these equations o- and T must be eliminated. We leave 
the work to the reader. There is obviously an equation of con- 
dition 

S.y(ft-a) = Q. 

Ex. 21. Classify the curves represented by an equation of 
the form 

a + xft + x s y 
P a + bx + cx* ' 

where a, ft, y are given vectors, and a, b, c given scalars, 

In the first place we remark that x 2 in the numerator merely 
adds a constant vector to the value of p, unless c = 0. 

Thus, if c do not vanish, the equation may be written, with 
a change of a and ft and in general a change of origin, 

a + xft 
a + bx + ex* ' 

and this again, by change of x and of a and ft, as 

a + xft 



It is obvious that this represents a plane curve. 

. , Sap a 2 + xSaft 

^3p = Xaft+xft*' 
T. Q. 12 



178 QUATERNIONS. [CHAP. IX. 

Hence both numerator and denominator of x are of the first 
degree in Sap, S(3p ; and therefore 


bap = 



car 
gives an equation of the third degree in p by the elimination of x. 

When we have Sa(3 = 0, 

a 2 



Sap = 



a + ex 



whence 



- , 

p /oap 

and a (Sap)* + c ^ (Spp}* = a* Sap, 

a conic section. 

If c = 0, then with a change of x, a, ft, y, the equation may be 
written 



a hyperbola so long at least as 6 does not also vanish. 

If 6 and c both vanish, the equation is obviously that of a 
parabola. 

If a and 6 both vanish, whilst c has a real value, we have 
again a parabola. 

If a vanish while 6 and c have real values, we have again 
a hyperbola. 

Ex. 22. Find the locus of a point at which a given finite 
straight line subtends a given angle. 



ART. 71.] FORMULAE AND THEIR APPLICATION. 179 

Take the middle point of the line as origin, and let a be the 
vectors of its ends. At p it subtends an. angle whose cosine is 



This, equated to a constant, gives the locus required. We 
may write the equation 



This is, obviously, a surface of the fourth order; a ring or 
tore formed by the rotation of a circle about a chord. When 
c = 0, i. e. when the angle is a right angle, the two sheets of this 
surface close up into the sphere 



A plane section (in the plane a, ft (suppose) where T(3 = 
and Sa(3 = 0) gives 

p = xa + yfi, 

(a* (1 - X s ) - yV} 2 = c 2 {(x - I) 2 + y 2 } {( + I) 2 + y 3 } a*, 
o r {1 - (x 2 + y 2 )} 2 - c 2 {( + y* + I) 2 - 4s 2 }, 

or, finally, 1 - (x* + /) = * -= , 



which, of course, denotes two equal circles intersecting at the 
ends of the fixed line. 

Ex. 23. A ray of light falls on a thin reflecting cylinder, shew 
that it is spread over a right cone. 

Let a be the ray, T a normal to the cylinder, p a reflected ray, 
/3 the axis of the cylinder. 

Then T is perpendicular to /?, or 

S(3r = Q .............................. (1). 

Again p and a make equal angles with T, on opposite sides of 
it, in one plane ; therefore 

p||TttT 

or V '. TOT/) = ........................... (2). 

122 



180 QUATERNIONS. [CHAP. IX. 

Eliminating T between (1) and (2) we have 



a 2 \Sap 
the equation of the right cone of which (B is the axis, and a a side. 



ADDITIONAL EXAMPLES TO CHAP. IX. 

1. Prove that S . (a + )8) (/3 + y) (y + a) = 2S . a/3y. 

2. S . Fa/3 F/3y Fya = - (Sa/3y) 2 . 

3. S . F ( Fa/3 F/3y) F ( F/3y Fya) F ( FyaFa) = -(S. a/3 7 ) 4 . 

4. ^ ( F/?y Fya) = y 2 ^a/3 - SfiySya. 



/3 2 (Sya) 



7. 

8. (a^y) 2 - o'jS V + 2ayS . a^y. 

9. ^ ( Fay Fj3yo Fyay8) = ISafiSpySyaS . afiy. 



10. The expression 

Fa/3 FyS + Fay F8^ + FaS F/3y 
denotes a vector. What vector 1 

(Tait's Quaternions. Miscellaneous Ex. 1.) 

1 1 . SapS . yS - SppS . ySa + SypS . Sa/3 - SSpS . a/3y = 0. 



12. (a^y) 2 = 2a 2 /3 2 y 2 + a 2 (/3y) 2 + /3 2 (ay) 2 + y 2 (a^) 2 - ay$aj3S(3y. 

(Hamilton, Elements, p. 346.) 



ART. 71.] FORMULA AND THEIR APPLICATION. 181 

13. With the notation of the Note, Art. 69. 5, we shall 
have 

DABC =OABC- OBCD + OCDA - ODAB. 

14. When A, B, (7, D are in the same plane, 

a.BCZ>-p.CDA+y.DAB-S.ABC = 0, 
where BCD, &c. are the areas of the triangles. 

15. SF. afiy + aV.ftyS + (3V. ySa + yF. Sa/? = 4S. a/fyS. 

16. Va/3 FyS + F/3y FSa + FyS FayS + VSa V(3y is a scalar. What 
is its geometrical meaning ? 

17. Find the equation of the sphere circumscribing a given 
tetrahedron. 

18. A straight line intersects a fixed line at right angles, and 
turns uniformly about it while it slides uniformly along it. Find 
the equation of the surface described (1) when the fixed line is 
straight, (2) when it is circular. 



CHAPTER X. 

VECTOR EQUATIONS OF THE FIRST DEGREE. 

WITH the object of giving the student an idea of one of the 
physical applications of Quaternions, we will treat the solution of 
linear and vector equations from an elementary kinematical point 
of view. For this purpose we choose the problem of the de- 
formation of a solid or fluid body, when all its parts are similarly 
and equally deformed. 

DBF. Homogeneous Strain is such that portions of a body, 
originally equal, similar, and similarly placed, remain after the 
strain equal, similar, and similarly placed. 

Thus straight lines remain straight lines, parallel lines remain 
parallel, equal parallel lines remain equal, planes remain planes, 
parallel planes remain parallel, and equal areas on parallel planes 
remain equal. Also the volumes of all portions of the body are 
increased or diminished in the same proportion, as is easily seen by 
supposing the body originally divided into small equal cubes by 
series of planes perpendicular to each other. After the strain, 
these cubes are all changed into similar, similarly placed, and 
equal parallelepipeds. 

It is thus obvious that a homogeneous strain is entirely deter- 
mined if we know into what vectors three given (non-coplanar) 
vectors are changed by it. Thus if a, ft, y become a', ft', y 



CHAP. X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 183 

respectively: any other vector, which may of course be expressed as 

p= * (aS 

is changed to 

1 ,M 

p = , Q- (a- *. 

b.aBy^ 

No needful generality is lost, while much simplification is 
gained, by taking a, B, y as unit vectors at right angles to one 
another. This is, in fact, the method already spoken of, i. e. the 
imaginary division of the body into small equal cubes, by three 
mutually perpendicular series of equidistant planes. We thus 
have 

p = - (aSap + BSBp + ySyp), 

p' = - (a' Sap + B'SBp + y'Syp), 

Comparing these expressions we see that Homogeneous Strain 
alters a vector into a definite linear and vector function of its 
original value. 

In abbreviated notation, we may write (as in Art. 63, though 
our symbol, as will soon be seen, is more general than that there 
employed) 

<f>p = (a Sap + B'SBp + y'Syp), 

where < itself depends upon nine independent constants involved 
in the three equations 

<f>a = a' I 

<t>y = y i 

For a', B', y may of course be expressed in terms of a, B, y : 
and, as they are quite independent of one another, the nine co- 
efficients in the following equations may have absolutely any 
values whatever ; 

<f>a = a Aa. + cB + b'y ] ^ 



<f>y = y = b a + a'B + Gy) 



184 QUATERNIONS. [CHAP. 

In discussing the particular form of <j!> which occurs in the 
treatment of central surfaces of the second order we found, Art. 44, 
that it possessed the property 

S . cr(j)p = S. p<f><r ......................... (&), 

whatever vectors are represented by p and o-. Remembering that 
a, (3, y form a rectangular unit system, we find from (a) 






with other similar pairs ; so that our new value of < satisfies (&) 
if, and only if, we have in (a) 



c = c 

The physical meaning of this condition, as will be seen im- 
mediately, is that the distortion expressed by < takes place without 
rotation. In this case the nine constants are reduced to six. 

But, although (6) is not generally true, we have 
S.<r<l>p = - (Sa'aSap 
= -S.p 

where the expression in brackets is a linear and vector function 
of o-, depending upon the same nine scalars as those in </> ; and 
which we may therefore express by < 7 , so that 

<}>'<T = -(aSa'<T + pSp'(r + ySy'<r) ............... (d). 

And with this we have obviously 

S . a-(f>p = S . p<'<r ......................... (e), 

which is the general relation, of which (6) is a mere particular 
case. 

By putting a, ft, y in succession for o- in (d) and referring to 
(a) we have 



tj/y = b'a + aft + Cy> 



X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 185 

Comparing (/) with (a) we see that 

$P = <f>'p> 

whatever be p, provided the conditions (c) be fulfilled. This agrees 
with the result already obtained. 

Either of the functions < and <', thus defined together, is 
called the Conjugate of the other : and when they are equal (i. e. 
when (c) is satisfied) < is called & Self-Conjugate function. As we 
employed it in Chap. VI, < was self-conjugate-; and, even had it 
not been so, it was involved (as we shall presently see) in such a 
manner that its non-conjugate part was necessarily absent. 

We may now write, as before, 

<j>p = ('a Sap + P'S/3p + y'Syp), 
and, by (d), 

<>'p = (aSa'p + PSft'p + ySy p). 

From these we have by subtraction, 
((/> <') p = <p <'p = aSa'p a'Sap + (3S{3'p fi'Sflp + y^j'p 

- V . Vaa'p + V. Vp/3'p + V . Vyy'p 

= 2F.ep .................................................... (y); 

if we agree to write 



We may now express that < is self-conjugate by writing 

e = 0, 

the physical interpretation of which equation is of the highest 
importance, as will soon appear. 

If we form by means of (a) the value of c as in (h) we get 
2c = (cy - 6'j8) + (ao - c'y) + (bj3 - a a) 



which obviously cannot vanish unless (as before) the three con- 
ditions (c) are satisfied. 



186 QUATERNIONS. [CHAP. 

By adding the values of <p and <'p above we obtain 
(< + <') p=<]>p + (j>'p = - (aSa'p + a'Sap +(3S(3'p +(3'Spp +ySy'p+y'Syp) 
= - F (apa' + (3pp + ypy') - p (Saa + Sftft' + Syy). 

As we have (by 69. 6) 

V . apa! = F . a'pa, &C. 
this new function of p is self-conjugate. 

This will easily be seen by putting < + <' for (p in (b) and re- 
membering that (by 69. 17) we have 

S . crapa = S . pa'cra = S . pacra', &c., &c. 

Hence we may write 

(< + (') p = 2arp ........................ (i), 

where the bar over OT signifies that it is self-conjugate, and the 
factor 2 is introduced for convenience. 

From (gr) and (i) we have 



,/ - r r | .......................... 

<p p 'ufp V f.p) 

If instead of <f>p in any of the above investigations we write 
(< + g) p, it is obvious that <j>p becomes (</>' + g)p: and the only 
change in the coefficients in (a) and (/") is the addition of g to 
each of the main series J, B, C. 

"We now come to Hamilton's grand proposition with regard to 
linear and vector functions. If < be such that, in general, the 
vectors 

p, <p, <f>"p 

(where < 2 p is an abbreviation for < (<p)) are not in one plane, then 
any fourth vector such as < 3 p (a contraction for < (<(<p))) can be 
expressed in terms of them as in 31. 5. 

Thus fj> 8 p = in 3 <j> 2 p m^p + mp .................. (&)> 

where m, m l , m 3 are scalars whose values will be found immedi- 
ately. That they are independent of p is obvious, for we may put 



X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 187 

a, ft, y in succession for p and thus obtain three equations of the 
form 

tj) 3 a = m 2 <f) 2 a m i (j)a + ma (I), 

from which their values can be found. For by repeated applica- 
tions of (a) we can express (I) in the form 

Aa + p + Cy = 0. 
This gives A = 0, $=0, C = 0. 

These are three equations connecting m, m^ m 2 , with the nine 
coefficients in (a). The other two groups of three equations, 
furnished by the other two equations of the form (?), are merely 
consistent with these ; and involve no farther limitations. This 
method, however, is very inferior to one which will shortly be 
given. 

Conversely, if quantities m, m l , m a can be found which satisfy 
(I), we may reproduce (&) by putting 

p = xa + yf$ + Zy 

and adding together the three expressions (I) multiplied by x, y, z 
respectively. For it is obvious from the expression for <f> that 

X<j>p e < (xp), X(f>*p = < 2 (xp), &C., 

whatever scalar be represented by x. 

If p, <p, and < 2 p are in the same plane, then applying the 
strain < again we find tj>p, (f> 2 p, < 3 p in one plane ; and thus equa- 
tion (k) holds for this case also. And it of course holds if <j>p is 
parallel to p, for then <jfp and <j> 3 p are also parallel to p. 

We will prove that scalars can be found which satisfy the 
three equations (F) (equivalent to nine scalar equations, of which, 
however, as we have seen, six depend upon the other three) by 
actually determining their values. 

The volume of the parallelepiped whose three conterminous 
edges are X, p., v is (31. 1) 

S . X.v. 



188 QUATERNIONS. 

After the strain its volume is 



so that the ratio 



[CHAP. 



S . 

S . 

,, . 

iO . A/JiV 



is the same whatever vectors X, //,, v may be ; and depends there- 
fore on the constants of </ alone. We may therefore assume 



and by inspection of (k) we find 

>v S . 
S . 



which gives the physical meaning of this constant in (/<;). As we 
may put if we please 



we see by (a) that 

S . (ba. 



m = 



S.afy 



A, c, b' 
c', B, a 
b, a', C 



which is the expression for the ratio in which the volume of each 
portion has been increased. This is unchanged by putting <' for 
</>, for it becomes, by (/), 

m - ' A, c', b 
c f , a' 
b', a, C 

Hence conjugate strains produce equal changes of volume. 
Recurring to (m) we may write it by (e) as 
S . A 



X.] VECTOR EQUATIONS OF T1IE FIRST DEGREE. 189 

from which, as X is absolutely any vector, we have 



or 



. =.^ , ^ 

V(j>fji<f>'v = m Vfj(.v) 



[In passing we may notice that (n) gives us the complete solution 
of a linear and vector equation such as 

<<r= 8, 

where 8 and <j* are given and cr is to be found. We have in fact 
only to take any two vectors //. and v which are perpendicular to 
8, and such that 

F/zv = S, 

and we have for the unknown vector 



<r = 

m 

which can be calculated, as < is given.] 

If in (n) we put < + g for <f> we must do so for the value of m 
in (m). Calling the latter N g we have 

S.(<j>+g)\ 



9* 



S . \fJLV 

S . Xp,<j)v + S . vXtfrfj. + S . fj.v(f)X 



/S . 



and by (n) (<f> + g} V (<' + g) n(<j>' + g) v = M,. V^v ......... (p), 



or 



/v 
'v) + g* F/*v] - M g 



From the latter of these equations it is obvious that 



must be a linear and vector function of F/xv, since all the other 
terms of the equation are such functions, 



190 QUATERNIONS. [CHAP. 

As practice in the use of these functions we will solve a 
problem of a little greater generality. The vectors 

Vpv, F</)'/AV, and V^'v 

are not generally coplanar. In terms of these (31. 5), let us 
express 

Let 

Operate by S . A, S . p, S . v successively, then 

S . fj.v(j)'X = xS . AJU.V + yS . vXfip. + zS . 
S . {j.v<f>'fji, yS '. v (*.$'[*-, 

S . [J.V(j>'v = zS . VfJ.(f>V. 

The two last equations give (by 69. 4) 

y = -l, = -!, 
and therefore the first gives 

$ . Jivt'X + S . v\(> ' JL 4- S . 



Hence, finally, 

<F/x.v = /i s F/iv V(f>' IJLV Vnfiv ............... (r). 

Substituting this in (q), and putting tr for F/X.V, which is any 
vector whatever, we have 

(< + 9) [<t>~* +ff(^- ^)+ff 2 ]^ = (m + P- 1 ff + ^ 2 ff i + Sf 3 ) <r, 
or, multiplying out, 

(m - g<f + ii a g<t> - g*<f> + gm^T l +g a <f t + g*^ + g 3 ) <r 



that is (- <jt s + p a <j> + m<l>~ 1 ) <r = /^o-, 

or (<#> 3 -fi 2 ^ a + /i 1 <)!)-m)o-=0. 

Comparing this with (k) we see that 

S . \u.d>v + S . vX<t>u. + S . u.vd>\ 
= = - y 

O . AfJLV 



i 1 

A})' . AylAV 

and thus the determination is complete. 



X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 191 

We may write (k), if we please, in the form 

m<p~ l p = m,p - m 2 <f>p + <p a p (&'), 

which gives another, and more direct, solution of the equation 
(above mentioned) 

(f)(T = 8. 

Physically, the result we have arrived at is the solution of 
the problem, " By adding together scalar multiples of any vector 
of a body, of the corresponding vector of the same strained homo-* 
geneously, and of that of the same twice over strained, to repre- 
sent the state of the body which would be produced by supposing 
the strain to be reversed or inverted." 

These properties of the function < are sufficient for many 
applications, of which we proceed to give a few. 

I. Homogeneous strain converts an originally spherical por- 
tion of a body into an ellipsoid. 

For if p be a radius of the sphere, tr the vector into which 
it is changed by the strain, we have 

o- = p, 

and Tp = C, 

from which we obtain 

jtyrvc, 

or tf.^-'o^-W-C", 

or, finally, 8 . ox^c/TV - - C\ 

This is the equation of a central surface of the second degree ; 
and, therefore, of course, from the nature of the problem, an 
ellipsoid. 

II. To find the vectors whose direction is unchanged by the 
strain. 

Here <p must be parallel to p or 
<t>p=gp. 
This gives <f> 2 p = g 2 p, &c., 



192 QUATERNIONS. [CHAP. 

so that by (k) we have 

g 3 - m 2 g 2 + m$ -m=0. 

This must have one real root, and may have three. Suppose g l to 
be a root, then 

<t>P ~ 9iP = > 
and therefore, whatever be A, 

S\({ip gfiXp 0, 
or S.p(^'\-g l \) = 0. 

Thus it appears that the operator <' g l cuts off from any vector 
A. the part which is parallel to the required value of p, and there- 
fore that we have 



where is absolutely any vector whatever. This may be written as 
(m 



t> 

^t/ 1 



The same result may more easily be obtained thus : 
The expression 

(< 3 - mtf + mrf -m)p = 0, 
being true for all vectors whatever, may be written 

(4>-<7 1 )(</>-<7 2 )(<-<7 3 )P = 0> 

and it is obvious that each of these factors deprives p of the por- 
tion corresponding to it : i. e. < g l applied to p cuts off the part 
parallel to the root of 

(< - grj o- = 0, &c., &c. 

so that the operator (<J> g y ) (<f> ~ g^) when applied to a vector 
leaves only that part of it which is parallel to or where 



X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 193 

III. Thus it appears that there is always one vector, and 
that there may be three vectors, whose direction is unchanged by 
the strain. 

DEF. Pure, or non-rotational, strain consists in altering the 
lengtlis of three lines at right angles to one anotJier, without altering 
tlieir directions. 

Hence if = 



the strain < is pure if, and not unless, p,, p 2 , p 3 form a rectangular 
system. [There is a qualification if two or more of ff l ff s g 3 be 
equal.] 

Hence, for a pure strain, we have 



and 

or S Pl <f>p 9 = S Pa fa. 

But we have, generally, 



As we have two other pairs of equations like these, we see 
that < = <' 

when the strain is pure. 

Conversely, if <f> = <j 

the three unchanging directions p l} p a) p s are perpendicular to one 
another. 

For, in this case, the roots of 

Jf,0 
are real. Let them be such that 

Cf-*)ft~0] 

(<*>-?,) p,=o[, 
(*-fc)iy-OJ 

T. Q. 13 



194- QUATERNIONS. [CHAP. 

tnen 



(because, by hypothesis, the strain is pure) 



for <t>Pz = 92P 2 and <p'p a = g a P a - 

Hence, except in the particular case of 

ffi = ff*> 
we must have 

8piP, = > 
whence the proposition. 

When <7, and g z are equal, p l and p 2 are each perpendicular 
to p a , but any vector in their plane satisfies 

<Jb(T gTjO- 0. 

When all three roots are equal, every vector satisfies 
<fxr - gp = 0. 

IV. Thus we see that when the strain is unaccompanied by 
rotation the three values of g are real. [But we must take care 
to notice that the converse does not hold. This will be discussed 
later.] If these values be real and different, there are three vectors 
at right angles to one another which are the only lines in the body 
whose directions remain unchanged. When two are equal, every 
vector parallel to a given plane, and all vectors perpendicular to 
it, are unchanged in direction. When all three are equal no 
vector has its direction changed. 

"V. There is, however, a peculiarity to be noticed, which dis- 
tinguishes true physical strain from the results of our mathe- 
matical analysis. When one or more of the values of g has a 
negative sign, we cannot interpret physically the result without 
introducing the idea of a pure strain which shall, as it were, pull 
the parts of an originally spherical portion of the body through 
the centre of the sphere, and so form an ellipsoid by turning a 
part of the body outside in. When two, only, are negative we 



X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 195 

can represent physically the result by introducing the conception 
of a rotation through two right angles about the third axis. But 
we began by assuming that there is no rotation ! Hence, for the 
case considered, all three roots must be positive. See end of next 
section (VI.). 

VI. This will appear more clearly if we take the case of a 
rigid body, for here we must have, whatever vectors be repre- 
sented by p and cr, 



Spcr = S . 

i. e. the lengths of vectors, and their inclinations to one another, 
are unaltered. In this case, therefore, the strain can be nothing 
but a rotation. It is easy to see that the second of these equa- 
tions includes the first; so that if, for variety, we take < as 
represented in equations (a), and write 

yft + zy, 



we have, for all values of the six scalars x, y, z, g, 77, , the follow- 
ing identity : 

'2 / O'2 '2 ^ 

+ (xr} + y) So! ft + (y + mi) Sfty' + ( + *) Sy'a. 
This necessitates 



i.e. the vectors a', ft', y form, like a, ft, y, a rectangular unit 
system. And it is evident that any and every such system 
satisfies the given conditions. But the system a', /8', y' must be 
similar to a, ft, y, i e. if a quadrant of positive rotation round a 
changes ft to y &c. a quadrant of positive rotation about a must 
change ft' to y' &c. 

When this is not the case, the system a, ft', y is the per- 

132 



196 QUATEENIONS. [CHAP. 

version of a, /?, y, i. e. its image in a plane mirror ; and the strain 
is impossible from a physical point of view. 

This is easily seen from another point of view. The volume 
of the parallelepiped whose edges are rectangular unit vectors 
a, (3, y is S . afiy 

if a positive quadrant of rotation round a brings (3 to coincide 
with y &c. But, in the perverted system, the volume has changed 
sign and is expressed by 

8.* fa. 

VII. It may be interesting to form, for this particular case, 
the equation giving the values of g. We have 
W _S.(<f> + g)a (< 
g " S.afiy 



S.afiy 



Recollecting that a, ft, y ; a', /?', y are systems of rectangular 
unit vectors, we find that this may be written 



Hence the roots of 

M g = 

are in this case ; first and always, 

?'=-*! 

which refers to the axis about which the rotation takes place 
secondly, the roots of 



Now the roots of this equation are imaginary so long as the 
coefficient of the first power of g lies between the limits * 2. 

Also the values of the several quantities W, S(3/3', Syy can 
never exceed the limits 1. When the system a, yS, y coincides 



X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 197 

with a', ft', y', the value of each of the scalars is 1, and the 
coefficient of the first power of g is + 2. When two of them are 
equal to + 1 and the third to - 1 we have the coefficient of the first 
power of g = 2. These are the only two cases in which the 
three values of g are all real. 

In the first, all three values of g are equal to 1, i. e. 

<1>P = P 

for all values of p, and there is no rotation whatever. In the 
second case there is a rotation through two right angles about 
the axis of the - 1 value of g. 

VIII. It is an exceedingly remarkable fact that, however a 
body may be homogeneously strained, there is always at least one 
vector whose direction remains unchanged. The proof is simply 
based on the fact that the strain-function depends on a cubic equa- 
tion (with real coefficients) which must have at least one real root. 

IX. As an illustration of what precedes (though one which 
must be approached cautiously), suppose a body to be strained- so 
that three vectors, a", (3", y" (not coplanar, and not necessarily 
at right angles to one another), preserve their direction, becoming 
e^", e a ft", e a y". Then we have 

<f> P S . a"ft"y" = e^'S. ft"y"p + e"S . y"a"p + e a y"S . a" ft" p. 
By the formulae (m, s) we have 



S (af'+PW 

= ~ 



S (a'P'ty" + ft"y"<}>a" + y'a'Aft") 

nravyr -= e ^ e * +e *> 

so that we have by (k) 

(^- 1 )(^- a )(4-?Jp=0. 

Though the values of g are here all real, we must not rashly 
adopt the conclusions of (iv.), for we must remember that a", j8", y" 
do not, like a, ft, y, necessarily form, a rectangular system. 



198 QUATERNIONS. [CHAP. 

In this case we have 

#pS . a"ft"y" = e r Vfi'fSa'p + e g Vy"a"Sft"p + eja." ft" Sy" p. 

So that, by (7i), 
2e . a."ft"y" = V. (e,a" Vp'y" + eft" Vy"a" + e.y" Va"ft") 



This vanishes, or the strain is pure, if either 

1. So." ft" = Sft"y" m Sy"a" = 0, 

Le. if a", ft", y" are rectangular, in which case e^ e 2 , e a may have 
any values ; or 

2. e l = e a = e 3 , in which case 

#pS. a"ft"y" = 6l { Vft"y"Sa"p+Vy"a"Sft"p + Va"ft"Sy"p} 

= e lP S.a"ft"y" by (69. 14), 
so that 

ftp = e t p = <f>p 

for every vector : a general uniform dilatation unaccompanied by 
change of direction. 

3. e l = e 2 , and a" and ft" both perpendicular to y". 

From what precedes it is evident that for the complete study 
of a strain we must endeavour to distinguish in each case between 
the pure strain and the merely rotational part. If a strain be 
capable of being decomposed into 1st a pure strain, 2nd a rotation, 
it is obvious that the vectors which in the altered state of the 
body become the axes of the strain- ellipsoid (i.) must have been 
originally at right angles to one another. 

The equation of the strain-ellipsoid is 



and in this it is obvious that (f>~* is self-conjugate, or at least is to 
be treated as such : for a non-conjugate term in <~ 2 /a would be (y) 
of the form Vep, 

and would therefore not appear in the equation. 



X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 199 

For the proper treatment of rotations, the following simple 
but excessively important proposition, due to Hamilton, forms the 
best starting-point. 

If q be any quaternion, the operator q ( ) q~ l turns the vector, 
quaternion, or body operated on round an axis perpendicular to the 
plane of q and through an angle equal to double that of q. 

For the proof we refer the reader to Hamilton's Lectures, 
282, Elements, 179 (1), or Tait, 353. It is obvious that the 
tensor of q may be taken to be unity, i. e. q may be considered as a 
mere versor, because the value of its tensor does not affect that of 
the operator. 

[A very simple but important example of this proposition is 
given by supposing q and r to be both vectors, a and fi let us say. 
Then 



is the result of turning /? conically through two right angles about 
a, i. e. if a be the normal to a reflecting surface and (3 the incident 
ray, a/3a -1 is the reflected ray.] 

Now let the strain <j> be effected by (1), a pure strain & (self- 
conjugate of course) followed by the rotation q ( ) q~\ We have, 
for all values of p, 



whence <p'p = 5r (q~ l pq). 

The interpretation is that, under the above definition, the con- 
jugate to any strain consists of the reversed rotation, followed by the 
pure strain. 

We may of course put, as in Chap, vi, 

vHp ejuSap + e^ftSfip + e 3 ySyp, 

where a, ft, y form a rectangular system. Hence 
<pp = e^aq^Sap + 



200 QUATERNIONS. [CHAP. 

Here the axes are parallel to 

qaq~\ q(3q~ l , qyq~ l , 
and we have 

S. qaq~*qpq~ l = S . qa(3q~ l = Sa{3 = 0, &c. 



So far the matter is nearly self-evident, but we now come to 
the important question of the separation of the pure strain from 
the rotation. By the formulae above we see that 



so that we have in symbols, for the determination of CT, the 
equation 

<f) <f) = W . 

That is, as we see at once from the statements above, any 
strain, followed by its conjugate, gives a pure strain, ivhich is the 
square (or the result of two applications) of the pure part of 
either. 

To solve this equation we employ expressions like (&). <ft'<f> 
being a known function, let us call it w, and form its equation as 

w 3 m 2 w 2 + nijto m = 0. 

Here the coefficients are perfectly determinate. 
Also suppose that the corresponding equation in OT is 

^-g^+g^-g^O, 

where g, g^ , g 2 are unknown scalars. By the help of the given 
relation TO* = w, 

we may modify this last equation as follows : 



whence 



= - *- 



X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 201 

i. e. tzr is given definitely in terms of the known function <o, as 
soon as the quantities g are found. But our given equation 



may now be written 



or w 3 - < - 2< a, 2 + - 2 o - = 0. 



As this is an equation between w and constants it must be 
equivalent to that already given : so that, comparing coefficients, 
we have 



9* = m; 

from which, by elimination of g and g s , we have 



The solution of the problem is therefpre reduced to that of this 
biquadratic equation ; for, when g l is found, g a is given linearly 
in terms of it. 

It is to be observed that in the operations above we have not 
been particular as to the arrangement of factors. This is due to 
the fact that any functions of the same operator are commutative 
in their application. 

Having thus found the pure part of the strain we have at once 
the rotation, for (v) gives 

^-'p^qpq- 1 , 
or, as it may more expressively be written, 



If instead of (v) we write 



202 



QUATERNIONS. 



[CHAP. 



we assume that the rotation takes place first, and is succeeded by 
the pure strain. This form gives 



and 

whence to is found as aboA'e. And then (v r ) gives 

5Tty = r( )r-\ 
Thus, to recapitulate, a strain < is equivalent to the pure 

strain *J <!>'<$> followed by the rotational strain <f> /^ , or to the 

' 



rotational strain -.- _ . < followed by the pure strain J <$>$'. 



This leads us, as an example, to find the condition that a given 
strain is rotational only, i.e. that a quaternion q can be found 
such that 



Here we have <' = q~ l ( ) q, 

or <' = <~ 1 

But m^)" 1 = T/IJ - m a <l> 

or mfi = ?, 

whose conjugate is m< = m, 7 

and the elimination of <' between these two equations gives 

+ < 2 ) + a 



= 



(m t m l mm^y + m 
(m a mm* + 2m 1 



jM^ + m*) 
(m 3 mm* + Zm l m i m) 4> 
+ (2m l + m* 



by using the expression for < 4 from the cubic in 



X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 203 

Now this last expression can be nothing else than the cubic 
in </> itself, else </> would have two different sets of constants in the 
form (&), which is absurd, as these constants, from, the mode in 
which they are determined, can have but single values. Thus we 
have, by comparing coefficients, 

m a a = 2m l + m 2 a - mm a m t 
mm. m 3 mm* + 2m.m., m 

| I * 

mm = m?rn mm + m* 



The first gives 

m l = mm a , 

by the help of which the second and third each become 

m 3 - m = 0. 
The value 

m = 

is to be rejected, as otherwise we should have been working with 
non-existent terms ; and m, as the ratio of the volumes of two 
tetrahedra, is positive, so that finally 

m 1, 
m 1 = m a , 
and the cubic for a rotational strain is, therefore, 



or > 

where m is left undetermined. 



By comparison with the result of (vn.) we see that in the 
notation there employed 



The student will perhaps here require to be reminded that 
in the section just referred to we employed the positive sign in 
operators such as < + g. In the one case the coefficients in the 
cubic are all positive, in the other they are alternately posi- 
tive and negative. The example we have given is a particularly 
valuable one, as it gives a glimpse of the extent to which the 



204 QUATERNIONS. [CHAP. 

separation of symbols can be safely carried in dealing with, these 
questions. 

DEP. A simple shear is a homogeneous strain in which all 
planes parallel to a fixed plane are displaced in the same direction 
parallel to that plane, and therefore through spaces proportional 
to their distances from that plane. 

Let a be normal to the plane, /? the direction of displacement, 
the former being considered as an unit- vector, and the tensor of 
the latter being the displacement of points at unit distance from 
the plane. 

We obviously have, by the definition, 

Sap = 0. 

Now if p be the vector of any point, drawn from an origin in 
the fixed plane, the distance of the point from the plane is 

Sap. 
Hence, if o- be the vector of the point after the shear, 



This gives 

<j>'p = p 

which may be written as 

= P -Tp.aS. 

so that the conjugate of a simple shear is another simple shear 
equal to the former. But the direction of displacement in each 
shear is perpendicular to the unaltered planes in the other. 

The equation for <f> is easily found (by calculating m, m 1} m a 
from (m), (s)) to be 

< 3 -3< B + 3e-l=0. 

Putting <'< = \J/, we easily find (with b = T/3) 
^ 3 _ (3 + b 2 ) ^ + (3 + b 2 ) $- 1 = 0. 
Solving by the process lately described, we find 



X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 205 

If b = 2, this gives ^ = 1, and the farther equation 

^ + ^'-13^-21 = 0, 
of which y l 3 is a root, so that 

*'-4r>-r-4 

and g v = I 2 J2. 

We leave to the student the selection (by trial) of the proper 
root, and the formation of the complete expressions for the pure 
and rotational parts of the strain in this simple and yet very 
interesting case. 

As a simple example of the case in which two of the roots of 
the cubic are unreal, take the vector function when the strain is 
equivalent to a rotation about the unit vector a ; the others of 
the rectangular system being /?, y. 

Here we have, obviously, 
<f>a = a, 
</? = (3 cos + y sin 9, 

<j>y = y cos - (3 sin 0, 
whence at once 

- <p = aSap + (ft cos + y sin 6) S(3p + (y cos (3 sin 6) Syp 
= (1 - cos &) aSap p cos 6 - Yap sin 0. 

Forming the quantities m, m l , m a as usual, we have 
< 3 - (1 + 2 cos 6} tf + (1 + 2 cos 0) < - 1 = 0, 
or (<-l)(< 2 -2cos0< + l) = 0, 

or (0 - 1) (< -cos - J^l sin 0} (< - cos 9 + J^l sin 6) = 0. 

Now 

- (< - 1) p = (1 - cos 0) (aSap + p)- sin & Fap, 

- (<f> - cos - J- 1 sin 6) p = (1 - cos 0} aSap + sin 6 (p A/-T- Yap), 

- (< - cos + J- 1 sin 6) p = (1 - cos 0) aSap - sin 6 (p J-l + 



206 QUATERNIONS. [CHAP. 

To detect the components which are destroyed by each of these 
factoi's separately, we have, by (n.), for (< 1), the vector 

(<t>*- 2 cos e < + 1) p = - ZaSap (1 - cos ff) ; 
so that (< 1) a = 0, 

which is, of course, true. Again 



which we leave to the student to verify. The imaginary directions 
which correspond to the unreal roots are thus, in this case, parallel 
to the Bivectors 



Here, however, we reach notions which, though by no means 
difficult, cannot well be called elementary. 

A very curious case, whose special interest however is rather 
mathematical than physical, is presented by the assumptions 



for then <f>p = ({3 + y) Sap + (y + a) Sfo + (a + ft) Syp 

l3 + y')p- (aSap 



where 8 is a known unit vector. This function is obviously self- 
conjugate. Its cubic is 

< 3 - 30 + 2 = = (< - I) 2 (< + 2), 

which might easily have been seen from the facts that 
1st, 08 = -2S, 
2nd, <a = a, if SaS = 0. 

The case is but slightly altered when the signs of a', /3', y are 
changed. Then 

<f>p = 3S$Sp p, 
and the cubic is 

< 3 - 3<- 2 = (< + I) 2 (<- 2) = 0. 



X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 207 

These are mere particular cases of extension parallel to the single 
axis 8. The general expression for such extension is obviously 

<}>p = p 
and we have for its cubic 



We will conclude our treatment of strains by solving the 
following problem : find the conditions which must be satisfied by 
a simple shear which is capable of reducing a given strain to a pure 
strain. 

Let < be the given strain, and let the shear be, as above, 
f*>l+/9&.a, 

then the resultant strain is 



Taking the conjugate and subtracting, we must have 
= i/<>- '' = <>-<j>' + S.<j>'a 



so that the requisite conditions are contained in the sole equation 

2e=V<l>'a(3. 
This gives (1) ./3e = 0, 

(2) S^'a(=0 = Sa4e. 

But (3) Sap (by the conditions of a shear), 

so that xa= V . fifa. 

Again, (4) 2e 2 = S . <'a p f = S.a 



or -ma=2V.p- l <l>f. 

Hence we may assume any vector perpendicular to e for /?, and 
a is immediately determined. 



208 QUATERNIONS. [CHAP 

When two of the roots of the cubic in < are imaginary let us 
suppose the three roots to be 



Let ft and y be such that 

* (ft + y J=l) = (e, + * 3 7 
Then it is obvious that, by changing throughout the sign of 
the imaginary quantity, we have 

< 08 - 7 J- 1 ) = (** ~ ** JPb (ft -7 N/^l ) 
These two equations, when expanded, unite in giving by 
equating the real and imaginary parts the values 



To find the values of a, /?, y we must, as before, operate on 
any vector by two of the factors of the cubic. 

As an example, take the very simple case . 

<f>p = e Vip. 

Here it is easily seen by (m), (s), that in = 0, m l = + e", m a = 0,- 
so that < 3 + e 2 < = 0, 

that is 

As operand take 



then 

\\ 

|| (_ jy _ fo + p ) 

II*. 
Again 



II - jy - Te + J - 1 (% - 
11,/y + &* - ^/"^T (; - ley). 



X.] VECTOR EQUATIONS OF THE FIRST DEGEEE. 209 

With a change of sign in the imaginary part, this will repre- 
sent 



so that /? =jy + kz, 

7 =jz - ky- 

Thus, as the student will easily find by trial, /3 and y form 
with a, a rectangular system. But for all that the system of 
principal vectors of </>, viz. 

a, (3yJ^l 

does not satisfy the conditions of rectangularity. In fact we see 
by the above values of /3 and y that 



It may be well to call the student's attention at this point to 
the fact that the tensors of these imaginary vectors vanish, for 



This gives a simple example of the new and very curious 
modifications which our results undergo when we pass to Si- 
vectors ; or, more generally, to Biquaternions. 

As a pendant to the last problem we may investigate the 
relation of two vector-functions whose successive application 
produces rotation merely. 

Here < = ^x~ l 

is such that by (w) 



or xx = l A = 

since each of these functions is evidently self-conjugate. This 
shews that the pure parts of the strains ty and x are the same, 
which is the sole condition. 

One solution is, obviously, 

X ' = x -', tf = r>, 
T. Q. 14 



210 QUATEKNTONS. [CHAP. 

i e. each of the two is itself a rotation ; and a new proof that any 
number of successive rotations can be compounded into a single 
one may easily be given from this. 

But we may also suppose either of i/r, x> suppose the latter, 
to be self-conjugate, so that 



or ur \I/ = Y , 

which leads to previous results. 

EXAMPLES TO CHAPTER X. 

1. If a, /?, y be a rectangular unit system 
S. Va<k<iVB<l>BVy<l>y = -mS. B^'-^ai, 



and therefore vanishes if <f> be self-conjugate. State in words the 
theorem expressed by its vanishing. 

2. With the same supposition find the values of 

SF. Fa<a. Vfi<j>{3 and of 2S . Fa<ctF/3</3. 
Also of 2 . aSa<j>a. 

3. When are two simple shears commutative ? 

4. Expand -^ - - in powers of <i. and reduce the result to 

1 e</> 

three terms by the cubic in <. 



5. Shew that *T. ^V = V ' 

P<PP<P P 



6. Why cannot we expand <' in terms of <, <f>, </> 21 ? 

7. Express Vp<f>p in terms of p, <^p, <f> 2 p, and from the result 
find the conditions that <p shall be parallel to p. 



X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 211 

8. Given the coefficients of the cubic in <, find those of the 
cubics in < 2 , </> 3 , &c. <^ n . 

9. Prove 



10. Ifm = 



.4, b, c 
a, B, c' 
a', b', C 



shew that M g = may be written as 



11. Interpret the invariants m 1 and m a in connexion with 
Homogeneous Strain. 

12. The cubics in tyr and \j/<f> are the same. 

13. Find the unknown strains < and ^ from the equations 



14. Shew that the value of F (<{>a-x a + 4>PxP + ^7X7) ^ ^ e 
same, whatever rectangular unit system is denoted by a, /?, y. 

15. Find a system of simple shears whose successive applica- 
tion results in a pure strain. 

16. Shew that, if < be self-conjugate, and , rj two vectors, 
the two following eqxiations are consequences one of the other : 



From either of them we obtain the equation : 

142 



212 QUATERNIONS. [CHAP. X. 

17. Shew that in general any self-conjugate linear and vector 
function may be expressed in terms of two given ones, the 
expression involving terms of the second order. 

Shew also that we may write 

< + z = a (ps + x) 3 + b (OT + x) (w + y) + c (w + yf t 
where a, b, c, x, y, z are scalars, and &, w the given functions. 
What character of generality is necessary in ta and o> ? How is 
the solution affected by non-self-conjugation in one or both 1 ? 

18. Solve the equations : 

(a) V.app=7.ayp, 

(b) 
(c) 

(d) apa~ l + Ppft' 1 = ypy 

(e) 



APPENDIX. 

WE have thought it would be acceptable to many students 
if we should give as an Appendix a brief, and in some cases 
even a detailed, solution of the most important and most difficult 
of the ADDITIONAL EXAMPLES. In doing so, we would add as 
a word of advice, that our solutions be employed simply for the 
purpose of comparison with those which shall occur to the student 
himself. 

CHAP. II. 

Ex.4. If AB = a, BC = (3, AP = ma, AP' = m'a, BQ=mf5, 
&c. ; then 

AE = AP + xPQ = AP' + x'P'Q 

gives ma. + x {(1 - m) a + m(3} = m'a + x' {(1 - ra') a + m'ft}, 
whence x = m' } and PE = m'PQ. 



Ex. 6. ABCD is a quadrilateral; AB = a, AC = /3, AD=y, 
AP = ma, BQ = m(fi- a), &c. 

The condition PQ + ES = 

gives (1 - m) a + m (ft a) + (1 ra) (y - ft) - my = 0, 

or (l-2ro)(a-j8 + y) = 0; 

an equation which is satisfied either when l-2m = 0, or when 
a-yS + y-0. 

The former solution is Ex. 5j the latter gives ABCD a 
parallelogram. 



214 QUATERNIONS. 

Ex. 10. Let a, b, c be the points in which the bisectors of 
the exterior angles at A, B, C meet the opposite sides. Let unit 
vectors along BC, CA, AB be a, (3,y; then with the usual nota- 
tion we have 



(1). 

Now Aa = x (/8 + y) = bfi + y (bj3 + cy) 

be 



~b-c 



&- 



, . be /0 
and Aa=i (p + y). 

A * v / ' 



Similarly 



a-6 v 

be 

therefore Ab = (3, (by 1), 

c-cr v 7 



Ac= y. 

a- b ' 

Hence (b c)Aa + (c - a) Ab + (a b)Ac = 0, 
and also ( b c) + (c a) + (a 6) = 0, 

therefore (Art. 13) a, b, c are in a straight line. 

COE. ba : ca :: b a : c-a. 

Ex. 12. If the figure of Ex. 11, Art. 23, be supposed to re- 
present a parallelepiped; then, with the notation of that example, 

the vector from to the middle point of OG is ^ (a + ft + 8), 
which is the same as the vector to the middle point of AF, viz. 



APPENDIX. 215 

Ex. 13. "With the figure and notation of Art. 31, the former 
part of the enunciation is proved by the equation 

1 a 



Also, if the edges AB, BO, CA be bisected in c, a, b, the mean 
point of the tetrahedron Oabc is evidently 



l/a + P 



y + a\ 
2 /' 



2 ~T 2 
which proves the latter part of the emmciation. 

Ex. 14. Here we have to do with nothing but the triangles 
on each side of OD. 

IfOQ = a, QA=pa, AP = p,PD = q{3; 



gives 



pq - 1 * 

Similarly, if OS = a, SB=p'a!, R 
T'0 = x'OD 
1 



gives x = 



p'q'-V 



But the data are - = , p = mq; hence 

pq=p'q', and x = x'; 
therefore T coincides with T. 



Ex. 15. 

we shall have, by making A0 = AP + PO = AR + RO, 



therefore p + q + r = 2. 

Ex. 1 7. Let RA = a, RB = p, AP = ma, AD=pa + qp; then 
PD =pa + qP~ ma, 



216 QUATERNIONS. 

and ES = P + PS=BQ + QS gives 

(1 + m) a + x (pa. + qfi- ma) = (1 + m) (3 + y (pa + q(3- m/3), 

1 + m 

whence x , 

in 

1 +m 1 + m . .. 

and JRS= (pa + qB) = - AD. 

m m 

[Or thus : 



QA =(l-m)a; 



CHAP. III. 

Ex. 5. Let ABGD be the quadrilateral; DA, DB, DC, a, (3, y 
respectively. 

a)+(y-a) = y (-a) + (- a)y 



Taking scalars, and applying 22. 3, there results, 



which is the proposition. 

Ex. 6. If a, /?, y be the vectors OA, OJB, OC corresponding 
to the edges a, b, c ; we have 



= abk + bci + caj, 
the negative square of which is the proposition given. 



APPENDIX. 217 

Ex. 7. If Sa(/3-y) = Q and Sj3(a-y) = Q, then, by sub- 
traction, will /Sy (a /?) = 0. 

Ex. 8. If a 2 = ((3 - y) 2 , P* = (y- a) 2 , / = (a - /3) 2 ; then will 



for these are the same equations in another form; and they prove 
that the corresponding vectors are at right angles to one another. 

Ex. 9. If OA, OB, 00, OD are a, (3, y, S; 

triangle DAB : DAG :: tetrahedron ODA B : ODAC 



:: triangle OAB : OAC, 

because the angles which S makes with the planes OAB, OAC are 
equal. 



CHAP. IY. 

Ex. 1. Let be the middle point of the common perpendi- 
cular to the two given lines ; a, a, the vectors from to those 
lines, unit vectors along which are ft, y ; p the vector to a point 
P in a line QR which joins the given lines j P being such that 
RP=mPQ; therefore 

p + a yy = m (a + xfi p). 

Now since a is perpendicular to both /? and y, the equation 
gives (l+m) Sap = (m - 1) a 2 a plane. 

Ex 2. Retaining what is necessary of the notation of the 
last example, let OS - 8. 

If PR perpendicular on y meet ft in Q, we have 
a + yy + RP = p, which gives yy 3 = Syp ; 
RQ = 2 a 4- xfi yy, which gives yy* xS{3y ; 



218 QUATEKNTONS. 

and SP a = e*PQ* gives 



which being of the second degree in p shews that the locus is a 
surface of the second order. See Chap. VI. 

Ex. 3. The equation of the plane is 

*Syp = a, 

which, being substituted in the equation of the surface, gives 
what is obviously the equation of a circle. 

Ex. 4. With the notation of Ex. 1, let 8, 8' be the perpen- 
diculars on the lines, 

then p+S = a + xp gives F/3S = - F/3 (p - a), 

and the condition given may be written 



Now (22. 9) 



whence p 2 - ZSap + a" + S*/3p = e* (p 2 + 2Sap + a 2 + S 2 y P ), 
a surface of the second order. 

Ex. 6. Sp (ft + y) = c, a plane perpendicular to the line which 
bisects the angle which parallels to the given lines drawn through 
make with one another. 

Ex. 7. a, (3 the vectors to the given points A, B, 

Syp = a, SSp = b 

the equations of the planes, y, 8 being unit vectors. 
xy, y8 the vector perpendiculars from A on the planes, then 
x = Say a, y Sa8 b, 

Sa(y + S)-(a+b) ................ (1). 



APPENDIX. 219 

Hence by the question 



or S(p-a)(y + S)=Q ........................ (2). 

Now equation (1) will give the sum of the perpendiculars on 
the planes from any other point in the line AB by simply writing 
a + z (/3 a) in place of a ; and from equation (2) this will pro- 
duce no change. 

Ex. 8. If /3' be the vector to C, equation (2) of the last 
example gives 



Now the sum of the perpendiculars from any other point in 
the plane will be found from equation (1) by writing 

a + z (/B - a) + z' (ft' - a) 
in place of a. Hence the proposition. 

Ex. 10. Tait's Quaternions, Art. 213. 

Ex. 11. Let a, (3, y, 8 be the vectors OA, OB, OC, OD 
then (34. 5, Cor.) 

8 - S. a/2y . (Fa/3 + F/?y + Fya)" 1 

abc (bci + caj + abk) /, * 

= (a*)* +{&$)* +()* ' 

Now 
triangle ABD : triangle ABC 

:: tetrahedron OABD : tetrahedron OABG 
: : S. a/38 : S. a/3y 
: : S . aJrijb : S . dbcijk 
:: (ab)* : (ab)> + (be} 3 + (ca)* 
: : (triangle AOB)* : (triangle ABC) 3 . 

(Chap. III., Additional Ex. 6.) 



220 QUATERNIONS. 

Ex. 12. This is merely the equation 

8 
p = at + ^, 

with t eliminated by taking the product of Vap, V(3p. (See 55. 3.) 

CHAP. V. 

Ex. 3. Let a, a' be the radii of the circles ', a, p the vectors 
from the centre of one of them to that of the other, and to the 
point whose locus is required ; then 



a a 

Ex. 7. This is the polar reciprocal of Ex. 3, Art. 40. 

Ex. 8. Let A be the origin, AB=B, AC = y, the vector to 
the centre a : then 

- V(AB . EC . CA) = V . (y - ft) y 

= y*(3-(?y 

= 2BSay - 2ySaB from the circle; 

,-. S.a7(AB.JBO.CA) = 0. 

Ex. 9. Tait, Art. 222. 

Ex. 10. Tait, Art. 221. 

Ex. 11. Tait, Art. 223. 

Ex. 12. Tait, Art. 232. 

CHAP. VI. 

Ex. 1. Let 8 be the vector to the given point, TT the vector to 
the point of bisection of a chord, B a vector parallel to the chord, 
all measured from the centre ; then 



(48); 



APPENDIX. 221 

from which, by making 



we get 8p<!>P 7 

an ellipse whose centre is at the point of bisection of the line 
which joins the given point with the centre of the given ellipse. 

Ex. 2. Let 26 be the shortest distance between the given 
lines ; their angle of inclination ; 2a the line of constant length ; 
then as in Ex. 2, Chap. IV., 



the former gives 

a 2 + 2/ 3 -2aycos0 = 4(a'-& 2 ) .................. (1), 

the latter 

4p = (* + y} (ft + y) + (x - y} (/? - y), 

which, since ft + y, ft y are vectors bisecting the angles between 
the lines and therefore at right angles to one another, is an equa- 
tion of the form of that in Art. 55. 2 ; whilst equation (1) satisfies 
the condition 



which is requisite for an ellipse. 

Ex. 3. Let a be a vector semi-diameter, parallel to a chord 
through ; 8 the vector to : then 

p = 8 + xa 

gives S<8 + 2xS8<f>a + x 2 Sa<f>a = 1, 

which, since >Sa<a=l, 

shews that the product of the two values of x is constant ; hence 
the rectangle by the segments of the chord varies as a*, which is 
the proposition. 



222 QUATERNIONS. 

Ex. 4. With the usual notation, let CE, CE' be semi- 
diameters parallel to DP, D'P, and let their vectors be ra (a (3), 
n (a + ft) ; then since P, D, E, E' are points in the ellipse, 



.'. 2m 2 =1. Similarly 2n s = 1, m = n, 
and DP : D'P :: !>-/?) : F(a + p) 

:: Tm(a-(3) : Tn(a+ ft) 
: : CE : CE'. 

COR. Since m = ~, CE : DP : : 1 : J2. 

v j 

Ex. 5. Put no.', np' in place of a, p in equation (1), Art. 43. 

Ex. 6, 7. With everything as in Ex. 4, CE, CE' being now 
semi-diameters in the direction of diagonals of the parallelogram, 



= 0; 
hence CE, CE' are conjugate. 

Ex. 8. S (a + ft) < (a + /3) = 2 gives an ellipse, whose equation 



is 

Sp4fp = l, where <'=|; 
hence the diameters of the locus are to those of the given ellipse 



Ex. 9. If y be a unit vector to which the lines are parallel, 
p, p' points in which the lines cut the ellipse, 



and "Sp^P = 1 gives 

2aSi<f>y + mSyijyy = 0| . 

Similarly 2bSj<}>y + nSy<j>y = 0) .................. ^ '' 



APPENDIX. 223 

Now Sp<j>p' = an Si<f>y + bmSj(f>y + mnSy$y 

= 0, by equations (1) ; 
.*. p, p are conjugate. 

COR. The same demonstration applies when the diameters 
from whose extremities parallels are drawn, are any conjugate 
diameters whatever, i, j being parallel to those diameters. 

Ex. 10. Let CP, CP 1 be any two semi-diameters, their vec- 
tors being a, a'; PQ the semi-ordinate to CP'; CQ na! ; then 

S (PQ . <j>a?) = 

gives S (a - na) <f>a? = 0, 

.*. n = /Satf>a. 

Now the area of the triangle QGP is proportional to 

V(CP.CQ), 
i. e. to n Vaa or to 

Sa<f>a . Faa', 

which, being symmetrical in a, a', proves the proposition. 

Ex. 11. If the tangent at P' meet CP produced in F, 

CT=ma; 
then, since PT is perpendicular to <j>af, 



-^r>> 

oa<pa 

and area P f CT is proportional to V(CP f .CT), i.e. to 
which is symmetrical in a, a'. 

Ex. 12. Let a, ft be the vector semi-diameters of the larger 
ellipse ; G the centre ; the centre of the smaller ellipse, whose 
equation is 

= c 



224 QUATERNIONS. 

y a vector along PQR ; then 



c 



_ 
- 



2 ' 



and since CQ = a + (3+ xy, 



hence PR is conjugate to CQ, and therefore bisected at Q. 
Ex. 13. This is simply a combination of 49. 2 and 49. 1. 

CHAP. VII. 

Ex. 3. The equation of the circle is 

9 



which by 52. 1 gives 

(a? Sap) 2 a'Sap = ^ a , 
ID 

2 

. '. Sap = r , 
4 

which (52. 11) is the proposition. 

Ex. 5. If be the centre of the circle, Q a point at which it 
meets the tangent at A ; then, with the notation of 55. 1, 



i. e. z* zy + ^- = 0, 
which gives two equal values of z ; hence the proposition. 



APPENDIX. 225 

Ex. 6. With any point as origin, let (3, y be the vectors to 
the two given points, TT the vector to the focus of one of the 
parabolas. Write aa in place of a in equation (1), Art. 52, a 
being a unit vector ; 

then - (p -)' = {a + Sa(p -v)}' .................... (1) 



whence, by subtraction, 

/3 2 - y 2 - 2S-H- (ft - y) = - Sa (J3 - y) {2a + Sa (ft - y) - 2Sair}, 

which gives a by a simple equation in TT; and then equation (1) 
becomes a quadratic in TT. 

Ex. 8. If two tangents meet at T, it is easy, as in Ex. 5, 
Art. 55, with the notation available for the focus, to find 



, a 

=- a H -- ^ p aa, 
4a 2 



and S(ST. ST') = will follow at once, from the fact that 



Ex. 9. Let P be the point of contact, PQ the chord, TEF the 
line parallel to the axis cutting the curve in Ej ; E the origin ; 





.,jr = 2 a -i- *p, J^J. = - ^ , 


F P Fl 


i pz> -K 1 -*' 2 '\o) . 


j JL J> / 


2/ a -^ 2 / J 




< tt' 


whence- 


z = - f r/ > y = - 



, 
t t 

.-. PF : FQ :: t : t' 

? tt?_ 
:: 2 : "2 

:: TE : EF. 

Ex. 10. This is evident from equation (1), Art. 52. 
T. Q. 15 



226 QUATERNIONS. 

Ex. 11. With the notation of Art. 52, let 



. '. x (a. 2p) = a + 8, 
x (a 2 - 2ap) - a 2 . 

But p, xp being vectors to the parabola, equation (1), Art. 
52, gives 



. '. X (a 2 Sap) = a 2 + OS/Sap, 
X (a 2 - 2Sap) = a 2 , 
.'. x=x', 
and the proposition is true (Euc. VI. 2). 

Ex. 14. Tait, Art. 43, Cor. 2. 

Ex. 15. 

CP= at + - gives CF= 2at, 

t 



so that the equation of RQPSf is 



whence for B and R' the values of x are 2 and 1 ; therefore 

C=3at, Ctf^l^, 

2> t 

QR=at-^ = PQ = 



APPENDIX. 227 

Ex. 1 6. If CR = aa ; a + m(3, a. - m(3 vectors parallel to the 
given conjugate diameters, 

CP = aa + x ( 



CD = aa. + SB' (a mjS) =at' ^ , 

tr 

give t = t'; therefore CP, CD are conjugate. 

Ex. 18. Adopting the figure and notation of Ex! 2 of the 
hyperbola, Art. 55, we have 



t 

therefore R 



- T) (to. - *-\ , 



and rQ. QR= (X s - Y 3 ) (to. - |Y 

= PO a , since ^ 8 -F 2 =l. 

As an example of combining not merely the forms but the 
results of the Cartesian Geometry with Quaternions, we will add 
one more example. 

If CP, CD; CP, CD 1 be two pairs of conjugate semi-diameters 
of an ellipse, PD' will be parallel to P'D. 

Let CP, CP be denoted, as in Art. 55. 2, by xa + yp, x'a + y r p 
respectively; then CD, CD' will be represented by 

b _ a , b ln 



with the conditions 

aY + 6V = a*b a , a*y" + b*x' 2 = a*b* ............ (1 ). 



Now vector D'P = (x + 1 y' j a 4- ( y - - x J /?, 



152 



228 QUATERNIONS. 



But equations (1) g r .ve, by subtraction, 

a . b , a , b 

x + 7 y : V x :: x + -rV ' V x 
b j a b y a 

therefore D'P is a multiple of DP' and consequently parallel to it. 
COR. PD' : P'D : : ay' + bx : ay + bx. 

CHAP. VIII. 

Ex. 1. With the notation of Additional Ex. 1, Chap. IV., 
the perpendiculars are 

p - a - xp, p + a - yy, 
so that Sfip = xft 2 , Syp = yy 2 ; 

and by the question, 

(p - a - p-^SppY = e* (p + a - y^Syp)*, 
a surface of the second order in p. 

Ex. 3. The equations Sp<f>p= 1, Sir<j>p = 1, with the condition 
Tr = X(f>p, give 

1 7T 2 

j STT(J>~ l ir = 1 , = 1 respectively, 

X X 



therefore Sir^T 1 -* = IT*, 

whence the Cartesian equation. 

Ex. 4. If a, p, y are the vector radii, 
Sa<f>a(SiU'a) 2 (SjUa) 2 



(Ta}*~ a 3 
&c. = &c, 

Adding and observing that Sa<f>a- 1, &c., there results 



1 1111 

= ~i + Ii + -2 ' 

a o c 



APPENDIX. 229 

Ex. 5. As in Ex. 8, Art. 64, 

<s? <*> 

and if vector OQ l = xfa, the ellipsoid gives 



Now ri = 

O*.OQ* x* 



and, since 



(Ex. 7, Art. 64), the result required is obtained by simply 
adding. 

Ex. 6. Let pk be the vector distance from the origin, of the 
plane parallel to xy, IT a point in it; then Sk (TT pk) = gives 
8-rrk = const. 

Now Spffrir = 1 is the equation of the plane of contact, and if 
zk be the point in which this plane cuts the axis of z, zSk<j>ir = 1, 
i.e. zSir<j>k = 1, gives z. 

Now tj>k is a multiple of k, and since Sirk is constant, z is 
constant. 

Ex. 7. The equations of the ellipsoids 

Sp<f>p = 1, S (p - a) < (p - a) = 1, 
give Sp(f>a = const, as the plane of contact. 

Ex. 8. If pa be the vector to the point in the line OA ; the 
equation of its polar plane is Spa<f>p = 1 ; and the square of the 
reciprocal of the perpendicular from the centre on this plane is 
2 . Hence the conclusion by Ex. 8, Art. 64. 



Ex. 9. Let p be the vector to P ; a, (3, y vector radii parallel 
to the chords ; then 

p + xa, p + y(3, p + zy, 



230 QUATERNIONS. 

will be the vectors to A, , C ; and since P, A, JB, C are 
points in the ellipsoid 

0, 2/S/Dj3 + y = 0, 

+ 2 = 0. 
The equation of the plane ABC is (34. 5) 

S . (TT p) (xya-P + yzfiy + zxya) = xyzS . a/2y, 
and since a, /?, y are at right angles to one another, 



therefore the equation of the plane ABC becomes 

<7 ( 



which is satisfied by 

TT p = m(f>p, 
where 



and therefore Ex. 4 above gives 

2 



CHAP. IX. 

Ex. 2 and 3. Employ formula 11. 
Ex. 5. Since 



formulae 4 and 6 give the required result. 
Ex. 6. Apply formula 10 to Ex. 5. 
Ex. 8. (a/3y) 2 = a/3y . a/3y = afiy (S.afty + V. a/?y) 



APPENDIX. 231 

Ex. 9. Formula 10 gives the vector of the product of three 
vectors a, /?, y, under the form a' /3' + y' where a = aSpy, &c. 

Hence the required scalar may be written 



and as the scalar part of this product is that which involves all of 
the three vectors a', /?', y we have exactly as in the demonstra- 
tion of formula 5, 



-a. 



a - 



', P, -y 

-', p 1 , y 



10. The scalar part, by formula 16, is reduced to 

SaSSpy - SaySpS - SaSSpy + SafiSyS + SaySfty - SapSyS, 
which is identically 0. 

The vector part, by formula 12, is 

aS. y&p-pS.yfa + aS. 8(3y-yS. Sfia + cuS. (3y8-SS. (3ya, 
which, by formula 13, reduces to 

2aS. 



12. If, for brevity, we denote S. a(3y, V . a(3y respectively by 
S and V, we have, by formula 7, 

2aj8Y + a 2 (Py)* + ft* (ay) 2 + y 2 (a/3)' - (a/3y) 2 
= 2aj8y . yj8a + (3ya . a/3y + ay@ . flay + a/?y . ya.fi - (afty) 2 



Y)(S-V+2ySap) - (S+ V) 3 



The student is recommended to verify a few examples such as 
the above; by putting 

a = i, P = ai + bj + ck, y = a'i + b'j + c'k, 



232 QUATERNIONS. 

with the conditions 

a* + b 3 + c a = l, a' 2 + 6' 2 + c' 2 = l. 

The quaternion equality will then reduce itself to four alge- 
braic equalities, one of which is obvious, and the others are 

p* + r a a' 2 a 2 + 2aa'm = 0, 
pq - mr + a'c' + ac- lac'm = 0, 
qr + mp + a'b' + ab 2ab'm - 0, 
where m = aa' + bb' + cc', p = ab' - a'b', 

q = be' b'c, r = ca c'a. 

Ex. 13. 

S. (a -8) (ft- 8) (y - 8) = S. aj8y - S . fa8 + S. yBa - S. Sap. 

Ex. 14. By 34. 8, we have 

aS.83 BCD 



therefore the same Article ives 



and since the scalar of the product of this vector by the vector 
perpendicular to the plane in which A, B, C, D lie gives the right- 
hand side of Ex. 13, we obtain 

a . BCD - ft . CDA + y . DAB - 8 . ABC = 0. 



CAMBRIDGE : PRINTED BY C. J. CLAY, M.A. AT THE UNIVERSITY PRESS. 



April 1892 



A Catalogue 

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For books of a less educational character on the subjects named below, see 
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CONTENTS 



GREEK AND LATIN 

CLASSICS- 
ELEMENTARY CLASSICS . 
CLASSICAL SERIES . 
CLASSICAL LIBRARY ; Texts, Com 

mentaries, Translations 
GRAMMAR, COMPOSITION, AND PHI 

LOLOGY 

ANTIQUITIES, ANCIENT HISTORY, 

AND PHILOSOPHY 

MODERN LANGUAGES AND 

LITERATURE- 
ENGLISH 

FRENCH 

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ITALIAN 

SPANISH 

MATHEMATICS 

ARITHMETIC 

BOOK-KEEPING 

ALGEBRA 

EUCLID AND PURE GEOMETR 

GEOMETRICAL DRAWING 

MENSURATION 

TRIGONOMETRY 

ANALYTICAL GEOMETRY 

PROBLEMS AND QUESTIONS IN HA 

THEMATICS .... 
HIGHER PURE MATHEMATICS 



MECHANICS 

PHYSICS 

ASTRONOMY 

HISTORICAL 

NATURAL SCDSNCES 

CHEMISTRY 

PHYSICAL GEOGRAPHY, GEOLOGY, 
AND MINERALOGY 

BIOLOGY 

MEDICINE 



PAGE 

26 

27 
29 
30 

30 

32 
32 
34 



HUMAN SCIENCES 

MENTAL AND MORAL PHILOSOPHY 35 

POLITICAL ECONOMY ... 36 

LAW AND POLITICS ... 37 

ANTHROPOLOGY .... 38 

EDUCATION 38 

TECHNICAL KNOWLEDGE- 
CIVIL AND MECHANICAL ENGINEER- 
ING 38 

MILITARY AND NAVAL SCIENCE . 39 

AGRICULTURE AND FORESTRY . 39 

DOMESTIC ECONOMY ... 40 

COMMERCE 40 

GEOGRAPHY 40 

HISTORY 41 

ART 43 

DIVINITY .... 44 



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